9.2 Statistical Inference for Two Population Means with Known Population Standard Deviations

A consumer advocacy group wants to study consumer satisfaction with their shopping experience at the country’s two biggest retailers.  The group surveyed consumers and asked them to rate one of the retailers in a number of different categories.  An overall satisfaction score out of 100 summarized the responses for each consumer sampled.  In a sample of 35 consumers for retailer A, the average overall satisfaction score was 79.  In a sample of 30 consumers for retailer B, the average overall satisfaction score was 71.  Based on prior experience with the satisfaction rating scale, the population standard deviation for retailer A is assumed to be 10 and the population standard deviation for retailer B is assumed to be 12.

1. Construct a 94% confidence interval for the difference in the mean satisfaction score for the two retailers.
2. Interpret the confidence interval found in part 1.
3. Is there evidence to suggest that the mean satisfaction score for retailer A is greater than the mean satisfaction score for retailer B?  Explain.

Solution:

1. Let retailer A be population 1 and retailer B be population 2.  These populations are independent because there is no relationship between the consumers sampled for each retailer.  From the question, we have the following information:
   

   Retailer A	Retailer B
   [latex]n_1=35[/latex]	[latex]n_2=30[/latex]
   [latex]\overline{x}_1=79[/latex]	[latex]\overline{x}_2=71[/latex]
   [latex]\sigma_1=10[/latex]	[latex]\sigma_2=12[/latex]

   The normal distribution applies because the sample sizes are both greater than or equal to 30.

   To find the confidence interval, we need to find the [latex]z[/latex]-score for the 94% confidence interval.  This means that we need to find the [latex]z[/latex]-score so that the entire area to the left of [latex]z[/latex] is [latex]\displaystyle{0.94+\frac{1-0.94}{2}=0.97}[/latex].

   

   Function	norm.s.inv	Answer
   Field 1	0.97	1.8807…

   So [latex]z=1.8807...[/latex]. The 94% confidence interval is

   [latex]\begin{eqnarray*} \\ \mbox{Lower Limit} & = & \overline{x}_1-\overline{x}_2-z \times \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}\\ & = & 79-71-1.8807... \times \sqrt{\frac{10^2}{35}+\frac{12^2}{30}} \\ & = & 2.796  \\ \\\mbox{Upper Limit} & = & \overline{x}_1-\overline{x}_2+z \times \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}\\ & = & 79-71+1.8807... \times \sqrt{\frac{10^2}{35}+\frac{12^2}{30}} \\ & = & 13.204 \\  \\ \end{eqnarray*}[/latex]

2. We are 94% confident that the difference in the mean satisfaction score for the two retailers is between 2.796 and 13.204.
3. Because 0 is outside the confidence interval and both limits are positive, it suggests that the difference in the means [latex]\displaystyle{\mu_1-\mu_2}[/latex] is greater than 0.  That is, [latex]\displaystyle{\mu_1-\mu_2 \gt 0}[/latex] ([latex]\mu_1 \gt \mu_2[/latex]).  This suggests that the mean for population 1 (retailer A) is greater than the mean for population 2 (retailer B).  So the mean satisfaction score for retailer A is greater than the mean satisfaction score for retailer B.

A floor cleaning company has been using Wax 1 to wax floors for a long time.  A new floor wax, Wax 2, has recently come on the market with the claim that it is longer lasting than Wax 1.  The company wants to investigate this claim.  The company waxed a sample of 20 floors with Wax 1 and found the average number of months the wax lasted was 2.7 months.  The company waxed a sample of 20 floors with Wax 2 and found the average number of months the wax lasted was 2.9 months.  Based on previous information, the standard deviation for the length of time Wax 1 lasts is 0.33 months and the standard deviation for the length of time Wax 2 lasts is 0.36 months.  Both populations have normal distributions.  At the 5% significance level, test if Wax 2 lasts longer, on average, than Wax 1.

Solution:

Let Wax 1 be population 1 and Wax 2 be population 2.  These populations are independent because there is no relationship between the length of time each type of wax lasts.  From the question, we have the following information:

Hypotheses:

[latex]\begin{eqnarray*} H_0: & & \mu_1-\mu_2=0 \\ H_a: & & \mu_1-\mu_2 \lt 0  \end{eqnarray*}[/latex]

p-value: 

This is a test on a the difference in two population means where the population standard deviation are known.  So we use a normal distribution to calculate the p-value.  Because the alternative hypothesis is a [latex]\lt[/latex], the p-value is the area in the left-tail of the distribution.

So the p-value[latex]=0.0335[/latex].

Conclusion:

Because p-value[latex]=0.0335 \lt 0.05=\alpha[/latex], we reject the null hypothesis in favour of the alternative hypothesis.  At the 5% significance level there is enough evidence to suggest that Wax 2 lasts longer than Wax 1.

A consumer advocacy group wants to compare the revolutions per minute (RPM) for two different engines.  The group believes that Engine A has a higher average RPM than Engine B.   In a sample of 40 Engine A’s, the sample mean number of RPMs was 1550.  In a sample of 30 Engine B’s, the sample mean number of RPMs was 1500.  Based on previous information, the standard deviation for the RPMs for Engine A is 75 and the standard deviation for Engine B is 65.  At the 1% significance level, is the average RPM for Engine A higher than for Engine B?

Solution:

Let Engine A be population 1 and Engine B be population 2.  These populations are independent because there is no relationship between the RPMs for the two engines.  From the questions, we have the following information:

Hypotheses:

[latex]\begin{eqnarray*} H_0: & & \mu_1-\mu_2=0 \\ H_a: & & \mu_1-\mu_2 \gt 0  \end{eqnarray*}[/latex]

p-value: 

This is a test on a the difference in two population means where the population standard deviation are known.  So we use a normal distribution to calculate the p-value.  Because the alternative hypothesis is a [latex]\gt[/latex], the p-value is the area in the right tail of the distribution.

So the p-value[latex]=0.0014[/latex].

Conclusion:

Because p-value[latex]=0.0014 \lt 0.01=\alpha[/latex], we reject the null hypothesis in favour of the alternative hypothesis.  At the 1% significance level there is enough evidence to suggest that the average RPM for Engine A is higher than for Engine B.

The student union at a local college owns two coffee shops on campus:  The Study Cafe and Coffee&Books.  The student union wants to find out if there is a difference the average amount students spend per transaction at each of the coffee shops.  In a sample of 65 transactions at the Study Cafe, the average amount spent was $9.40.  In a sample of 50 transactions at Coffee&Books, the average amount spent was $10.15.  Based on previous information, the standard deviation for the amount spent at the Study Cafe is $1.35 and the standard deviation for Coffee&Books B is $2.70.  At the 5% significance level, is there a difference in the average amount spent per transaction at the two coffee shops?

Solution:

Let the Study Cafe be population 1 and Coffee&Books be population 2.  These populations are independent because there is no relationship between the amount spent at each coffee shop.  From the question, we have the following information:

Hypotheses:

[latex]\begin{eqnarray*} H_0: & & \mu_1-\mu_2=0 \\ H_a: & & \mu_1-\mu_2 \neq 0  \end{eqnarray*}[/latex]

p-value: 

This is a test on a the difference in two population means where the population standard deviation are known.  So we use a normal distribution to calculate the p-value.  Because the alternative hypothesis is a [latex]\neq[/latex], the p-value is the sum of the area in the two tails of the distribution.

We need to know if the sample information relates to the left or right tail because that will determine how we calculate out the area of that tail using the normal distribution.  In this case, the [latex]\overline{x}_1 \lt \overline{x}_2[/latex] ([latex]9.4 \lt 10.15[/latex]), so the sample information relates to the left tail of the normal distribution.  This means that we will calculate out the area in the left tail using norm.dist.  However, this is a two-tailed test where the p-value is the sum of the area in the two tails and the area in the left tail is only one half of the p-value.  The area in the left tail equals the area in the right tail and the p-value is the sum of these two areas.

So the area in the left tail is 0.0360, which means [latex]\frac{1}{2}[/latex](p-value)[latex]=0.0360[/latex].  This is also the area in the right tail, so

p-value[latex]=0.0360+0.0360=0.0720[/latex]

Conclusion:

Because p-value[latex]=0.0720 \gt 0.05=\alpha[/latex], we do not reject the null hypothesis.  At the 5% significance level there is not enough evidence to suggest that there is a difference in the average amount spent at the two coffee shops.