Hypothesis Tests

$ (a.) \\[3ex] Two-tailed\:\:Hypothesis\:\:Test \\[3ex] p-value = 0.0888 \\[3ex] \alpha = 5\% = \dfrac{5}{100} = 0.05 \\[5ex] 0.0888 \gt 0.05 \\[3ex] p-value \gt \alpha \\[3ex] $ Decision: Fail to reject the null hypothesis

$ (b.) \\[3ex] Two-tailed\:\:Hypothesis\:\:Test \\[3ex] p-value = 0.0888 \\[3ex] \alpha = 10\% = \dfrac{10}{100} = 0.1 \\[5ex] 0.0888 \le 0.1 \\[3ex] p-value \le \alpha \\[3ex] $ Decision: Reject the null hypothesis

$ (c.) \\[3ex] One-tail\:(Right-tailed)\:\:Hypothesis\:\:Test \\[3ex] p-value = 0.0888 \\[3ex] \alpha = 5\% = \dfrac{5}{100} = 0.05 \\[5ex] 0.0888 \gt 0.05 \\[3ex] p-value \gt \alpha \\[3ex] $ Decision: Fail to reject the null hypothesis

One-Sample Proportion

$ H_0:\;\; p = 97\% = 0.97 \\[3ex] q = 1 - p = 1 - 0.97 = 0.03 \\[5ex] n = 1217 \\[3ex] \hat{p} = 89\% = 0.89 \\[3ex] z = \dfrac{\hat{p} - p}{\sqrt{\dfrac{pq}{n}}} \\[9ex] z = \dfrac{0.89 - 0.97}{\sqrt{\dfrac{0.97(0.03)}{1217}}} \\[9ex] z = -16.36020652 \\[3ex] z \approx -16.36 $

One-Sample Proportion

p = 30% = 0.3
(a.) The null and alternative hypotheses are:

$ H_0:\;\; p = 0.3 \\[3ex] H_1:\;\; p \ne 0.3 \\[3ex] $ (b.) The test statistic, z = 0.6172
(c.) The probability value, p = 0.5371

One-Sample Mean

$ H_0:\;\; \mu = 124\;mmHg \\[3ex] n = 279 \\[3ex] \bar{x} = 123.93\;mmHg \\[3ex] s = 15.71\;mmHg \\[3ex] t = \dfrac{\bar{x} - \mu}{\dfrac{s}{\sqrt{n}}} \\[9ex] t = \dfrac{123.93 - 124}{\dfrac{15.71}{\sqrt{279}}} \\[9ex] t = -0.0744258763 \\[3ex] t \approx -0.07 $

One-Sample Proportion

p = 32% = 0.32
(a.) The null and alternative hypotheses are:

$ H_0:\;\; p = 0.32 \\[3ex] H_1:\;\; p \lt 0.32 \\[3ex] $ (b.) The test statistic, z = -1.9174
(c.) The probability value, p = 0.0276

One-Sample Standard Deviation

$ H_0:\;\; \sigma = 10\;bpm \\[3ex] H_1:\;\; \sigma \lt 10\;bpm \\[5ex] n = 174 \\[3ex] s = 8.6bpm \\[3ex] \chi^2 = \dfrac{s^2(n - 1)}{\sigma^2} \\[5ex] \chi^2 = \dfrac{8.6^2(174 - 1)}{10^2} \\[5ex] \chi^2 = 127.9508 \\[3ex] \chi^2 \approx 127.95 $

One-Sample Proportion

$ H_0:\;\; p = 0.57 \\[3ex] H_1:\;\; p \gt 0.57 \\[5ex] (a.) \\[3ex] x = 115 \\[3ex] n = 200 \\[3ex] \hat{p} = \dfrac{x}{n} = \dfrac{115}{200} = 0.575 \\[5ex] (b.) \\[3ex] p_0 = 0.57 \\[3ex] (c.) \\[3ex] p = p_0 = 0.57 \\[3ex] q = 1 - p \\[3ex] q = 1 - 0.57 \\[3ex] q = 0.43 \\[3ex] z = \dfrac{\hat{p} - p}{\sqrt{\dfrac{pq}{n}}} \\[9ex] z = \dfrac{0.575 - 0.57}{\sqrt{\dfrac{0.57 * 0.43}{200}}} \\[9ex] z = 0.1428279973 \\[3ex] z \approx 0.14 \\[3ex] $



The value of the test statistic tells that the observed proportion of students who passed introductory chemistry was 0.14 standard errors above the null hypothesis proportion of students.

0.1% = 10%
The​ p-value is the probability that if the null hypothesis is​ true (the person is guessing), a test statistic (p) will have a value as extreme as or more extreme than the observed value (gets 14 or more right).
So, the correct option is: D.: The probability that a person will get 14 or more​ right, if the person is truly​ guessing, is about 10​%.

One-Sample Proportion
We are only interested in the number of adults who regularly use supplemental vitamins

$ H_0:\:\: p = 0.47 \\[3ex] H_1:\:\: p \ne 0.47 \\[3ex] x = 91 \\[3ex] n = 170 \\[3ex] \hat{p} = \dfrac{x}{n} \\[5ex] \hat{p} = \dfrac{91}{170} \\[5ex] \hat{p} = 0.5352941176 \\[3ex] p = 0.47 \\[3ex] q = 1 - p \\[3ex] q = 1 - 0.47 \\[3ex] q = 0.53 \\[3ex] z = \dfrac{\hat{p} - p}{\sqrt{\dfrac{pq}{n}}} \\[9ex] z = \dfrac{0.5352941176 - 0.47}{\sqrt{\dfrac{0.47 * 0.53}{170}}} \\[9ex] z = \dfrac{0.06529411765}{\sqrt{0.001465294118}} \\[5ex] z = \dfrac{0.06529411765}{0.03827916036} \\[5ex] z = 1.705735367 \\[3ex] \alpha = 5\% = \dfrac{5}{100} = 0.05 \\[5ex] \dfrac{\alpha}{2} = \dfrac{0.05}{2} = 0.025 \\[5ex] \underline{Classical\:\:Approach} \\[3ex] -z_{\dfrac{\alpha}{2}} = -1.9599639861189817 \\[5ex] z_{\dfrac{\alpha}{2}} = 1.9599639861189817 \\[5ex] -z_{\dfrac{\alpha}{2}} \lt z \lt z_{\dfrac{\alpha}{2}} \\[5ex] -1.96 \lt 1.71 \lt 1.96 \\[3ex] $ It does not fall in the critical region
Decision: Do not reject the null hypothesis

$ \underline{P-value\:\:Approach} \\[3ex] P(z) = P(z \lt -1.705735367) + P(z \gt 1.705735367) \\[3ex] P(z) = 0.04402868083136646 + 0.04402868083136646 \\[3ex] P(z) = 0.08805736166 \\[3ex] P(z) \gt \alpha \\[3ex] 0.0881 \gt 0.05 \\[3ex] $ Decision: Do not reject the null hypothesis

Conclusion: There is insufficient evidence to warrant the rejection of the claim that forty seven percent of the adult population regularly uses supplemental vitamins.

(f.)
To change the conclusion, there should be sufficient evidence to warrant the rejection of the claim that forty seven percent of the adult population regularly uses supplemental vitamins.
This means that there should sufficient evidence to warrant the rejection of the null hypothesis.
This means that we have to change the decision as well (the decision led to the conclusion)
This means that we have to reject the null hypothesis
The easiest way to do this is to use the P-value Approach
For Two-tailed tests, we reject the null hypothesis if $P-value \le \alpha$
This means that we have to find a significance level that is greater than or equal to $P(z)$
$P(z) = 0.08805736166$
The common significance level that is greater than $0.08805736166$ is $0.1$
$0.1 = 10\%$
Therefore, the significance level that will change the conclusion is $10\%$

Let us verify with the calculator.

Student: Are you saying that if we test the hypothesis using a $1\%$ level of significance, the conclusion will be the same?
Teacher: That is correct.
Of the three most common significant levels, only the 10% level of significance will give a different conclusion.

One-Sample Proportion

$ H_0:\:\: p = 0.45 \\[3ex] H_1:\:\: p \gt 0.45 \\[3ex] x = 194 \\[3ex] n = 360 \\[3ex] \hat{p} = \dfrac{x}{n} \\[5ex] \hat{p} = \dfrac{194}{360} \\[5ex] \hat{p} = 0.5388888889 \\[3ex] p = 0.45 \\[3ex] q = 1 - p \\[3ex] q = 1 - 0.45 \\[3ex] q = 0.55 \\[3ex] z = \dfrac{\hat{p} - p}{\sqrt{\dfrac{pq}{n}}} \\[9ex] z = \dfrac{0.5388888889 - 0.45}{\sqrt{\dfrac{0.45 * 0.55}{360}}} \\[9ex] z = \dfrac{0.08888888889}{\sqrt{0.0006875}} \\[5ex] z = \dfrac{0.08888888889}{0.0262202212} \\[5ex] z = 3.390089206 \\[3ex] \alpha = 5\% = \dfrac{5}{100} = 0.05 \\[5ex] \dfrac{\alpha}{2} = \dfrac{0.05}{2} = 0.025 \\[5ex] \underline{Classical\:\:Approach} \\[3ex] z_{\dfrac{\alpha}{2}} = 1.9599639861189817 \\[5ex] z \gt z_{\dfrac{\alpha}{2}} \\[5ex] 3.39 \gt 1.96 \\[3ex] $ It falls in the critical region
Decision: Reject the null hypothesis

$ \underline{P-value\:\:Approach} \\[3ex] P(z \gt 3.390089206) = 0.000349349418287237 \\[3ex] P(z \gt 3.390089206) \le \alpha \\[3ex] 0.0003 \le 0.05 \\[3ex] $ Decision: Reject the null hypothesis

Conclusion: There is sufficient evidence to support the claim that the initiative is working to increase retention.

One-Sample Proportion

$ (a.) \\[3ex] x = 39 \\[3ex] n = 100 \\[3ex] \hat{p} = \dfrac{x}{n} = \dfrac{39}{100} = 0.39 \\[5ex] (b.) \\[3ex] p_0 = 0.34 \\[3ex] (c.) \\[3ex] p_0 = p = 0.34 \\[3ex] q = 1 - p = 1 - 0.34 = 0.66 \\[3ex] z = \dfrac{\hat{p} - p}{\sqrt{\dfrac{pq}{n}}} \\[9ex] z = \dfrac{0.39 - 0.34}{\sqrt{\dfrac{0.34 * 0.66}{100}}} \\[9ex] z = 1.055500827 \\[3ex] z \approx 1.06 \\[3ex] $ The value of the test statistic tells that the observed proportion of prisoners was 1.06 standard errors above the null hypothesis proportion of prisoners.

(a.) 72% = 0.72
Null hypotheses: The passing rate for introductory chemistry is 72​%.
Alternative hypotheses: A college chemistry instructor thinks the use of embedded tutors will improve the success rate in introductory chemistry courses.

$ H_0:\;\; p = 0.72 \\[3ex] H_1:\;\; p \gt 0.72 \\[3ex] $ (b.) The p-value is the probability (0.0091) that if the null hypothesis is​ true (the passing rate for introductory chemistry is 72​%), a test statistic (z = 2.36) will have a value as extreme as or more extreme than the observed value (will greatly improve the success rate in introductory chemistry courses).
In the context of the question, this implies that:
The probability of 159 or more introductory chemistry students passing out of a random sample of 200 students is 0.0091​, assuming the population proportion is 0.72.

(c.) In this​ situation, the p-value is ​small (less than the significance level: 0.05), so reject the null hypotheses.
The instructor should believe the success rate has improved.

Two-Samples Proportion

$\hat{p_1}$ is the first sample proportion (proportion of urban families who regularly eat lunch together)
$\hat{q_1}$ is the complement of the first sample proportion
$\hat{p_2}$ is the second sample proportion (proportion of rural families who regularly eat lunch together)
$\hat{q_2}$ is the complement of the second sample proportion
$\overline{p}$ is the pooled sample proportion
$\overline{q}$ is the complement of the pooled sample proportion

$ H_0:\:\: p_1 = p_2 \:\:OR\:\: p_1 - p_2 = 0 \\[3ex] H_1:\:\: p_1 \lt p_2 \\[3ex] \hat{p_1} = \dfrac{x_1}{n_1} \\[5ex] \hat{p_1} = \dfrac{12}{75} \\[5ex] \hat{p_1} = 0.16 \\[3ex] \hat{p_2} = \dfrac{x_2}{n_2} \\[5ex] \hat{p_2} = \dfrac{40}{140} \\[5ex] \hat{p_2} = 0.2857142857 \\[3ex] 0.16 \lt 0.2857142857 \\[3ex] $ Yes, these sample proportions show what the researcher expected.
The sample proportion of urban families who regularly eat lunch together is lower than the sample proportion of rural families who regularly eat lunch together because $0.16 \lt 0.2857142857$

$ \overline{p} = \dfrac{x_1 + x_2}{n_1 + n_2} \\[5ex] \overline{p} = \dfrac{12 + 40}{75 + 140} \\[5ex] \overline{p} = \dfrac{52}{215} \\[5ex] \overline{p} = 0.2418604651 \\[3ex] \overline{q} = 1 - 0.2418604651 \\[3ex] \overline{q} = 0.7581395349 \\[3ex] z = \dfrac{(\hat{p_1} - \hat{p_2}) - (p_1 - p_2)}{\sqrt{\dfrac{\overline{p} * \overline{q}}{n_1} + \dfrac{\overline{p} * \overline{q}}{n_2}}} \\[10ex] z = \dfrac{(0.16 - 0.2857142857) - 0}{\sqrt{\dfrac{0.2418604651 * 0.7581395349}{75} + \dfrac{0.2418604651 * 0.7581395349}{140}}} \\[10ex] z = \dfrac{-0.1257142857}{\sqrt{\dfrac{0.1833639805}{75} + \dfrac{0.1833639805}{140}}} \\[10ex] z = \dfrac{-0.1257142857}{\sqrt{0.002444853074 + 0.001309742718}} \\[7ex] z = \dfrac{-0.1257142857}{\sqrt{0.003754595792}} \\[7ex] z = \dfrac{-0.1257142857}{0.06127475656} \\[7ex] z = -2.05164888 \\[3ex] \alpha = 5\% = \dfrac{5}{100} = 0.05 \\[5ex] \dfrac{\alpha}{2} = \dfrac{0.05}{2} = 0.025 \\[5ex] \underline{Classical\:\:Approach} \\[3ex] -z_{\dfrac{\alpha}{2}} = -1.9599639861189817 \\[5ex] z \lt -z_{\dfrac{\alpha}{2}} \\[5ex] -2.05164888 \lt -1.9599639861189817 \\[3ex] $ It falls in the critical region
Decision: Reject the null hypothesis

$ \underline{P-value\:\:Approach} \\[3ex] P(z) = P(z \lt -2.05164888) P(z \lt -2.05164888) = 0.020101899077793972 \\[3ex] P(z \lt -2.05164888) \le \alpha \\[3ex] 0.020101899077793972 \le 0.05 \\[3ex] $ Decision: Reject the null hypothesis

Conclusion: There is sufficient evidence to support the claim that the sample proportion of urban families who regularly eat lunch together is lower than the sample proportion of rural families who regularly eat lunch together.

(g.)

$ Test\:\:statistic:\:\: z = -2.05164888 \approx -2.05 \\[3ex] P-value:\:\: P(z) = 0.020101899077793972 \approx 2\% $

(a.) If the population proportion, p is not given, then assume p = 50% = 0.5
50% of 30 trials = 0.5(30) = 15
The person should get 15 trials right.

$ (b.) \\[3ex] \underline{Person\;A} \\[3ex] x = 21 \\[3ex] n = 30 \\[3ex] \hat{p} = \dfrac{x}{n} = \dfrac{21}{30} = 0.7 \\[5ex] \underline{Person\;B} \\[3ex] x = 19 \\[3ex] n = 30 \\[3ex] \hat{p} = \dfrac{x}{n} = \dfrac{19}{30} = 0.6333333333 \\[5ex] $ 0.7 > 0.6333333333; therefore
Person A will have a smaller p-value because that person's number of successes is further from the hypothesized number of successes

One-Sample Mean

$ \underline{1st\;\;Step:\;\;Hypothesis} \\[3ex] H_0:\;\; \mu = 98.6 \\[3ex] H_1:\;\; \mu \ne 98.6 \\[5ex] \underline{2nd\;\;Step:\;\;Test\;\;Statistic} \\[3ex] Sample: \\[3ex] n = 10 \\[3ex] 10 \lt 30 ...use\;\;t-statistic \\[3ex] t = \dfrac{\bar{x} - \mu}{\dfrac{s}{\sqrt{n}}} \\[9ex] \bar{x} = 98.09 \\[3ex] s = 0.9480389115 \\[3ex] \implies \\[3ex] t = \dfrac{98.09 - 98.6}{\dfrac{0.9480389115}{\sqrt{10}}} \\[9ex] t = -1.701155498 \\[5ex] \underline{3rd\;\;Step:\;\;Critical\;\;value\;\;of\;\;t} \\[3ex] df = n - 1 \\[3ex] df = 10 - 1 \\[3ex] df = 9 \\[3ex] Two-tails \\[3ex] \alpha = 0.05 \\[3ex] Critical\;\;t = \pm 2.262 \\[5ex] \underline{4th\;\;Step:\;\;Classical\;\;Approach} \\[3ex] -2.262 \lt -1.701155498 \\[3ex] Critical\;\;t \;\lt\; Test\;\;Statistic \;\;(for\;\;negative\;\;values) \\[3ex] $ This implies that the test statistic is not in the critical region
Decision: Do not reject the null hypothesis

5th Step: Probability Value
By Technology:







$ p-value = 0.12312501 \\[3ex] \alpha = 0.05 \\[3ex] 0.12312501 \gt 0.05 \\[3ex] p-value \;\gt\; significance\;\;level \\[3ex] $ 6th Step: Probability Value Approach
The probability value is greater than the significance level
Decision: Do not reject the null hypothesis

Conclusion: There is insufficient evidence to conclude that the population mean is not 98.6 °F at a significance level of 0.05.

Student: Can you find the probability value for t test by Tables?
Is technology the only way to find it?
Teacher: We can use the t distribution table to find it.
We need to look for the exact value of the test statistic for 9 degrees of freedom and find the significance level that corresponds to that exact value.
If we do not find the exact value, then we have to interpolate between two closest in-between values of significance levels.
Let's do it.

Let p be the probability value that corresponds to the test statistic of -1.701155498 at 9 degrees of freedom
This is a two-tailed test so it goes both ways.
Because the test statistic is negative, we choose a negative value that corresponds to that value.



$ \underline{Linear\;\;Interpolation} \\[3ex] \dfrac{p - 0.1}{0.2 - 0.1} = \dfrac{-1.701155498 - (-1.833)}{-1.383 - (-1.833)} \\[5ex] \dfrac{p - 0.1}{0.1} = \dfrac{-1.701155498 + 1.833}{-1.383 + 1.833} \\[5ex] \dfrac{p - 0.1}{0.1} = \dfrac{0.131844502}{0.45} \\[5ex] p - 0.1 = \dfrac{0.1(0.131844502)}{0.45} \\[5ex] p - 0.1 = 0.0292987782 \\[3ex] p = 0.0292987782 + 0.1 \\[3ex] p = 0.1292987782 \\[3ex] $ Student: But that is not the value we got using technology
Teacher: Well, yes...but if we rounded both values to 2 decimal places (technology and interpolated values), they are the same
Student: I guess the technology value is more accurate...
Teacher: That is correct.
The values from the t distribution table from which we interpolate (depending on the closest in-between values of interpolation), gives the approximate value to few number of decimal places.

(a.) Number of African Americans the researcher would expect in a jury pool of 200 people is:

$ 24\% \;\;of\;\; 200 \\[3ex] = 0.24(200) \\[3ex] = 48\;\;African\;\;Americans \\[3ex] $ (b.) Expected hypothesized number of successes: 48
Pool A number of succcesses: 35 African American
Pool B number of successes: 26 African American

Which of them is further from 48?

|48 − 35| = |13| = 13
|48 − 26| = |22| = 22
13 > 12
Pool B will have a smaller p-value because that pool's number of African American people is further from the hypothesized number.

Goodness-of-Fit test
Goodness-of-Fit test is always a right-tailed test

Null hypotheses: H_0: All days of the week have an equal chance of being selected.
Alternative hypotheses: H_1: At least one day of the week has a different chance of being selected.

Test statistic, χ² = 1731.386
Critical χ² = 12.592

Critical Value Method
1731.386 > 12.592
Test statistic, χ² > Critical χ²
This implies that the Test statistic, χ² is in the critical region
Decision: Reject the null hypotheses.

Probability Value Method
0.0000 < 0.05
P-value is < Significance level
Decision: Reject the null hypotheses.

Conclusion: There is sufficient evidence to warrant rejection of the claim that the days of the week are selected with a uniform distribution.
Interpretation: It does not appear that all days have the same chance of being selected.

One-Sample Mean

$ (a.) \\[3ex] \underline{Hypothesis} \\[3ex] H_0:\;\; \mu = 25 \\[3ex] H_1:\;\; \mu \gt 25 \\[5ex] (b.) \\[3ex] t = 2.93 \\[5ex] (c.) \\[3ex] p-value = 0.003 \\[3ex] $ (d.) Since the​ p-value is less than or equal to the significance level, reject H₀
There is sufficient evidence to conclude that the mean BMI is more than 25 using a significance level of 0.05.

One-Sample Proportion

$ (a.) \\[3ex] x = 14 \\[3ex] n = 80 \\[3ex] \hat{p} = \dfrac{x}{n} = \dfrac{14}{80} = 0.175 \\[5ex] $ (b.) 18% = 0.18
Null hypotheses: Nationally the readmission rate for patients with pneumonia is 18​%.
Alternative hypotheses: A hospital was interested in knowing whether their readmission rate for pneumonia was less than the national percentage.

$ H_0:\;\; p = 0.18 \\[3ex] H_1:\;\; p \lt 0.18 \\[3ex] (c.) \\[3ex] q = 1 - p = 1 - 0.18 = 0.82 \\[3ex] z = \dfrac{\hat{p} - p}{\sqrt{\dfrac{pq}{n}}} \\[9ex] z = \dfrac{0.175 - 0.18}{\sqrt{\dfrac{0.18 * 0.82}{80}}} \\[9ex] z = -0.1164050493 \\[3ex] z \approx -0.12 \\[3ex] $ This implies that:
The value of the test statistic tells that the observed proportion of readmissions was 0.12 standard errors below the null hypothesis proportion of readmissions. (it is below because of the negative sign)

(d.) The​ p-value is the probability (0.45) that if the null hypothesis is​ true (Nationally the readmission rate for patients with pneumonia is 18​%.), a test statistic (z = -0.12) will have a value as extreme as or more extreme than the observed value (their readmission rate for pneumonia was very less than the national percentage).
In the context of the question, this implies that:
The probability of getting 14 or fewer readmissions for pneumonia of a random sample of 80 patients with pneumonia is 0.45, assuming the population proportion is 0.18.
In this​ situation, the​ p-value is not small (because it is greater than 0.05), which indicates that the null hypothesis should not be doubted.

One-Sample Mean

$ (a.) \\[3ex] \underline{Hypothesis} \\[3ex] H_0:\;\; \mu = 200 \\[3ex] H_1:\;\; \mu \gt 200 \\[5ex] (b.) \\[3ex] t = 1.21 \\[5ex] (c.) \\[3ex] p-value = 0.116 \\[3ex] $ (d.) Since the​ p-value is greater than the significance level, do not reject H₀
There is insufficient evidence to conclude that the mean cholesterol level is more than 200 ​mg/dL at a significance level of 0.05.

1st Step: Hypothesis
Null Hypothesis: H_0: Humans are like random number generators and produce numbers in equal quantities.
Alternative Hypothesis: H_1: Humans are not like random number generators and do not produce numbers in equal quantities.

Each Observed Value = Times Chosen (Given in the Table)
Total Number of Observed Values = 43
Each Expected Value = roughly equal numbers of all five integers = $\dfrac{43}{5} = 8.6$

2nd Step: Chi-Square Test Statistic

Integer Selection

Integer	Observed Values, O	Expected Values, E	$O - E$	$(O - E)^2$	$\dfrac{(O - E)^2}{E}$ (to 5 decimal places)
One	16	8.6	7.4	54.76	6.36744
Two	2	8.6	−6.6	43.56	5.06512
Three	7	8.6	−1.6	2.56	0.29767
Four	11	8.6	2.4	5.76	0.66977
Five	7	8.6	−1.6	2.56	0.29767
$\chi^2 = \Sigma \dfrac{(O - E)^2}{E} = 12.69767$

Technology Approach: 1st Method:

$ L_1 = Observed \\[3ex] L_2 = Expected \\[3ex] L_3 = L_1 - L_2 \\[3ex] L_4 = L_3^2 \\[3ex] L_5 = L_4 \div L_2 \\[3ex] $

OR

Technology Approach: 2nd Method:
Degrees of Freedom = 5 − 1 = 4





3rd Step: Critical Value Method
Critical value of χ² at 4 degrees of freedom and 0.05 significance level = 9.48772
12.69767 > 9.48772
Test statistic χ² > Critical value χ²
This implies that the Test statistic χ² is in the critical region.
Decision: Reject the null hypothesis.

4th Step: Probability Value Method
By technology: probability = 0.0128514953
0.0128514953 < 0.05
Probability value is less than the level of significance
Decision: Reject the null hypothesis.

Conclusion: Humans have been shown to be different from random number generators.

Student: Can you find the probability value for GOF test by Tables?
Is technology the only way to find it?
Teacher: We can use the χ² table to find it.
We need to look for the exact value of the test statistic for 4 degrees of freedom and find the significance level that corresponds to that exact value.
If we do not find the exact value, then we have to interpolate between two closest in-between values of significance levels.
Let's do it.

Let p be the probability value that corresponds to the test statistic of 12.69767 at 4 degrees of freedom



$ \underline{Linear\;\;Interpolation} \\[3ex] \dfrac{p - 0.015}{0.01 - 0.015} = \dfrac{12.69767 - 12.3391}{13.2767 - 12.3391} \\[5ex] \dfrac{p - 0.015}{-0.005} = \dfrac{0.35857}{0.9376} \\[5ex] p - 0.015 = \dfrac{-0.005(0.35857)}{0.9376} \\[5ex] p - 0.015 = -0.0019121694 \\[3ex] p = -0.0019121694 + 0.015 \\[3ex] p = 0.0130878306 \\[3ex] $ Student: But that is not the value we got using technology
Teacher: Well, yes...but if we rounded both values to 3 decimal places (technology and interpolated values), they are the same
Student: I guess the technology value is more accurate...
Teacher: That is correct.
The values from the Chi-square table from which we interpolate (depending on the closest in-between values of interpolation), gives the approximate value to few number of decimal places.

(a.) The null hypothesis is p = 0.20, where p is the probability of a correct answer.
The null hypotheses is true (or assumed to be true)
It makes sense that the probability of getting a correct answer by guessing is 20% so we assume the null hypothesis to be true
The test statistic compares the observed outcome with the outcome of the null hypothesis (expected outcome).
For this case, the observed outcome is close to the expected outcome.
Hence, the z-test statistic will be close to 0.

(b.) Because the null hypothesis is true, obtaining an unusual result is not likely.
Hence, the p-value will not be small.

(a.) μ = 20
Confidence Interval: I am 95​% confident the population mean is between 20.9 and 22.3 pounds.
20 is not in-beween 20.9 and 22.3
It is not included. It is less than the lower limit of the confidence interval.
Confidence Interval Method: Reject the​ hypothesis, because the interval does not contain the proposed mean.

$ (b.) \\[3ex] \underline{1st\;\;Step:\;\;Hypothesis} \\[3ex] H_0:\;\; \mu = 20 \\[3ex] H_1:\;\; \mu \ne 20 \\[3ex] $ (c.) 2nd Step: Test Statistic
Sample size, n = 4
4 is less than 30
Therefore, we shall use the t test







t test statistic = 7

$ \underline{3rd\;\;Step:\;\;Critical\;\;t-value} \\[3ex] \alpha = 0.05 \\[3ex] df = n - 1 \\[3ex] df = 4 - 1 \\[3ex] df = 3 \\[3ex] $ Critical t-value at 0.05 significance level and 3 degrees of freedom = ± 3.182 (two-tailed test)
Because the test statistic is positive:
Critical t-value = 3.182

4th Step: Critical Value Method
7 > 3.182
test statistic > critical value of t
This implies that the test statistic is in the critical region.
Decision: Reject the null hypothesis.

5th Step: Probability Value
p-value = 0.0059862557

5th Step: Probability Value Method
0.0059862557 < 0.05
probability value is less than the significance level
Decision: Reject the null hypothesis.