9.5 Using Technology for Hypothesis Testing with Two Samples

1. two means
2. unknown
3. Student's t
4. [latex]{\overline{X}}_{1}–{\overline{X}}_{2}[/latex]
5. [latex]H_0: \mu_1 = \mu_2[/latex] Null hypothesis: the means of the final exam scores are equal for the online and face-to-face statistics classes.
   [latex]H_a: \mu_1 \lt \mu_2[/latex] Alternative hypothesis: the mean of the final exam scores of the online class is less than the mean of the final exam scores of the face-to-face class.
6. left-tailed
7. [latex]\text{p-value} = 0.0011[/latex]
8. Reject the null hypothesis
9. The professor was correct. The evidence shows that the mean of the final exam scores for the online class is lower than that of the face-to-face class. At the 5% level of significance, from the sample data, there is sufficient evidence to conclude that the mean of the final exam scores for the online class is less than the mean of final exam scores of the face-to-face class.

An image of the distribution with the shaded p-value is shown below.

This is a test of two independent groups, two population means, population standard deviations known.

Random Variable: [latex]{\overline{X}}_{1} - {\overline{X}}_{2}[/latex] = difference in the mean number of months the competing floor waxes last.

[latex]H_0: \mu_1 \le \mu_2[/latex]

[latex]H_a: \mu_1 > \mu_2[/latex]

The words "is more effective" says that wax 1 lasts longer than wax 2, on average. "Longer" is a “>” symbol and goes into [latex]H_a[/latex]. Therefore, this is a right-tailed test.

Distribution for the test: The population standard deviations are known so the distribution is normal. Using the formula, the distribution is:

[latex]{\overline{X}}_{1} - {\overline{X}}_{2} \sim N\left(0,\sqrt{\frac{{0.33}^{2}}{20}+\frac{{0.36}^{2}}{20}}\right)[/latex]

Since [latex]\mu_1 \le \mu_2[/latex] then [latex]\mu_1 - \mu_2 \le 0[/latex] and the mean for the normal distribution is zero.

Calculate the p-value using the normal distribution: [latex]\text{p-value} = 0.1799[/latex]

Graph:

 

[latex]{\overline{X}}_{1} - {\overline{X}}_{2} = 3 - 2.9 = 0.1[/latex]

Compare [latex]\alpha[/latex] and the p-value: [latex]\alpha = 0.05[/latex] and [latex]\text{p-value} = 0.1799[/latex]. Therefore, [latex]\alpha \lt \text{p-value}[/latex].

Make a decision: Since [latex]\alpha \lt \text{p-value}[/latex], do not reject [latex]H_0[/latex].

Conclusion: At the 5% level of significance, from the sample data, there is not
sufficient evidence to conclude that the mean time wax 1 lasts is longer (wax 1 is more effective) than the mean time wax 2 lasts.

Two types of medication for hives are being tested to determine if there is a difference in the proportions of adult patient reactions. Twenty out of a random sample of 200 adults given medication A still had hives 30 minutes after taking the medication. Twelve out of another random sample of 200 adults given medication B still had hives 30 minutes after taking the medication. Test at a 1% level of significance.

Solution

The problem asks for a difference in proportions, making it a test of two proportions.

Let A and B be the subscripts for medication A and medication B, respectively. Then pA and pB are the desired population proportions.

Random Variable: [latex]P^{\prime}_A - P^{\prime}_B =[/latex] difference in the proportions of adult patients who did not react after 30 minutes to medication A and to medication B.

[latex]H_0: p_A = p_B[/latex], so [latex]p_A - p_B = 0[/latex]

[latex]H_a: p_A \neq p_B[/latex], so [latex]p_A - p_B \neq 0[/latex]

The words "is a difference" tell you the test is two-tailed.

Distribution for the test: Since this is a test of two binomial population proportions, the distribution is normal:

[latex]{p}_{c}=\frac{{x}_{A}+{x}_{B}}{{n}_{A}+{n}_{B}}=\frac{20+12}{200+200}=0.08[/latex], and [latex]1–{p}_{c}=0.92[/latex]

[latex]{{P}^{\prime }}_{A}–{{P}^{\prime }}_{B}~N\left[0,\sqrt{\left(0.08\right)\left(0.92\right)\left(\frac{1}{200}+\frac{1}{200}\right)}\right][/latex]

[latex]P^{\prime}_A – P^{\prime}_B[/latex] follows an approximate normal distribution.

Calculate the p-value using the normal distribution: [latex]\text{p-value} = 0.1404[/latex].

Estimated proportion for group A: [latex]{{p}^{\prime }}_{A}=\frac{{x}_{A}}{{n}_{A}}=\frac{20}{200}=0.1[/latex]

Estimated proportion for group B: [latex]{{p}^{\prime }}_{B}=\frac{{x}_{B}}{{n}_{B}}=\frac{12}{200}=0.06[/latex]

Graph:

[latex]P^{\prime}_A – P^{\prime}_B = 0.1 – 0.06 = 0.04[/latex]

Half the p-value is below –0.04, and half is above 0.04.

Compare [latex]\alpha[/latex] and the p-value: [latex]\alpha = 0.01[/latex] and the [latex]\text{p-value} = 0.1404[/latex, so [latex]\alpha \lt \text{p-value}[/latex].

Make a decision: Since [latex]\alpha \lt \text{p-value}[/latex], do not reject [latex]H_0[/latex].

Conclusion: At a 1% level of significance, from the sample data, there is not sufficient evidence to conclude that there is a difference in the proportions of adult patients who did not react after 30 minutes to medication A and medication B.

Solution

This is a test of two population proportions. Let M and F be the subscripts for males and females. Then [latex]p_M[/latex] and [latex]p_F[/latex] are the desired population proportions.

Random variable: [latex]p^{\prime}_F − p^{\prime}_M =[/latex] difference in the proportions of males and females who sent “sexts.”

[latex]H_0: p_F = p_M[/latex], so [latex]H_0: p_F – p_M = 0[/latex]

[latex]H_a: p_F \lt p_M[/latex], so [latex]H_a: p_F – p_M \lt 0[/latex]

The words "less than" tell you the test is left-tailed.

Distribution for the test: Since this is a test of two population proportions, the distribution is normal:

[latex]{p}_{c}=\frac{{x}_{F}+{x}_{M}}{{n}_{F}+{n}_{M}}=\frac{156+183}{2169+2231}=\text{0}\text{.077}[/latex], so [latex]1-{p}_{c}=0.923[/latex]

Therefore, [latex]{{p}^{\prime }}_{F}–{{p}^{\prime }}_{M}\sim N\left(0,\sqrt{\left(0.077\right)\left(0.923\right)\left(\frac{1}{2169}+\frac{1}{2231}\right)}\right)[/latex]

[latex]p^{\prime}_F – p^{\prime}_M[/latex] follows an approximate normal distribution.

Calculate the p-value using the normal distribution:

[latex]\text{p-value} = 0.1045[/latex]

Estimated proportion for females: 0.0719

Estimated proportion for males: 0.082

Graph:

 

[latex]H_a: p_W > p_A[/latex], so [latex]H_a: p_W – p_A > 0[/latex]

The words "more popular" indicate that the test is right-tailed.

Distribution for the test: The distribution is approximately normal:

[latex]{p}_{c}=\frac{{x}_{W}+{x}_{A}}{{n}_{W}+{n}_{A}}=\frac{134+12}{1343+232}=0.0927[/latex]

[latex]1-{p}_{c}=0.9073[/latex]

Therefore,

[latex]{{p}^{\prime}}_{W}–{{p}^{\prime}}_{A}\sim N\left(0,\sqrt{\left(0.0927\right)\left(0.9073\right)\left(\frac{1}{1343}+\frac{1}{232}\right)}\right)[/latex]

[latex]{{p}^{\prime }}_{W}–{{p}^{\prime }}_{A}[/latex] follows an approximate normal distribution.

Calculate the p-value using the normal distribution:

[latex]\text{p-value} = 0.0077[/latex]

Estimated proportion for group A: 0.10

Estimated proportion for group B: 0.05

Graph:

Decision: Since [latex]\alpha > \text{p-value}[/latex], reject the [latex]H_0[/latex].

Conclusion: At the 5% level of significance, from the sample data, there is sufficient evidence to conclude that a larger proportion of white cell phone owners use iPhones than African Americans.

Solution

Corresponding "before" and "after" values form matched pairs. (Calculate "after" – "before.")

The data for the test are the differences: {0.2, –4.1, –1.6, –1.8, –3.2, –2, –2.9, –9.6}

The sample mean and sample standard deviation of the differences are [latex]\overline{{x}_{d}}=–3.13[/latex] and [latex]{s}_{d}=2.91[/latex]. Verify these values.

Let [latex]{\mu }_{d}[/latex] be the population mean for the differences. We use the subscript [latex]d[/latex] to denote "differences."

Random variable: [latex]{\overline{X}}_{d}[/latex] = the mean difference of the sensory measurements

[latex]H_0: \mu_d \ge 0[/latex]

The null hypothesis is zero or positive, meaning that there is the same or more pain felt after hypnotism. That means the subject shows no improvement. μd is the population mean of the differences.

[latex]H_a: \mu_d \lt 0[/latex]

The alternative hypothesis is negative, meaning there is less pain felt after hypnotism. That means the subject shows improvement. The score should be lower after hypnotism, so the difference ought to be negative to indicate improvement.

Distribution for the test: The distribution is a Student's t with [latex]df = n – 1 = 8 – 1 = 7[/latex]. Use [latex]t_7[/latex]. (Notice that the test is for a single population mean.)

Calculate the p-value using the Student's t-distribution: [latex]\text{p-value} = 0.0095[/latex]

Graph:

[latex]{\overline{X}}_{d}[/latex] is the random variable for the differences.

The sample mean and sample standard deviation of the differences are:

[latex]{\overline{x}}_{d} = -3.13[/latex]

[latex]{\overline{s}}_{d} = 2.9[/latex]

Compare α and the p-value: [latex]\alpha = 0.05[/latex] and [latex]\text{p-value} = 0.0095[/latex], so [latex]\alpha > \text{p-value}[/latex].

Make a decision: Since [latex]\alpha > \text{p-value}[/latex], reject [latex]H_0[/latex]. This means that [latex]\mu_d \lt 0[/latex] and there is improvement.

Conclusion: At a 5% level of significance, from the sample data, there is sufficient evidence to conclude that the sensory measurements, on average, are lower after hypnotism. Hypnotism appears to be effective in reducing pain.