Question

The following relationship holds between hypothesis tests and confidence intervals for one-mean t-procedures: For a two-tailed hypothesis test at the significance level $\alpha$, the null hypothesis $H_0:\mu ={\mu }_0$will be rejected in favor of the alternative hypothesis $H_a:\mu >{\mu }_0$if and only if ${\mu }_0$lies outside the $\left(1-\alpha \right)$-level confidence interval for $\mu$. In each case, illustrate the preceding relationship by obtaining the appropriate one-mean t-interval and comparing the result to the conclusion of the hypothesis test in the specified exercise.

Part (a): Exercise 9.113

Part (b): Exercise 9.116

Short answer

Part (a): Both conclusions are same.

This means that the conclusion for confidence interval is same as the conclusion for hypothesis test.

Part (b): Both conclusions are same.

This means, that the conclusion for confidence interval is same as the conclusion for hypothesis test.

Step-by-step solution

Part (a) Step 1. Given information.

Consider the given question,

The null hypothesis is $H_0:\mu ={\mu }_0$.

The alternative hypothesis is$H_a:\mu >{\mu }_0$.

Part (a) Step 2. Compute the confidence interval.

On using the MINITAB procedure,

1. Choose Stat>Basic Statistics>1-Sample t.
2. In Summarized data, enter the sample size 20 and mean 4.760.
3. In Standard deviation, enter the value 2.297.
4. In Perform hypothesis test, enter the test mean as 4.55.
5. Check Options, enter Confidence level as 90.
6. Choose not equal is alternative.
7. Click OK in all dialogue boxes.

Hence, from the MINITAB output, the90% confidence interval is$\left(3.872,5.648\right)$.

Part (a) Step 3. Write the conclusion.

The population mean 4.55, which lies between lower and upper limit. Therefore, the null hypothesis is not rejected at 10% level.

It can be concluded that the data provide sufficient evidence to conclude that the amount of television watched per day lasy year by average person differed from that in 2005.

Using the Hypothesis test,

The null hypothesis is given below,

$H_0:\mu =4.55$

The average person do not watched 4.55 hours of television per day in 2005.

The alternative hypothesis is given below,

$H_a:\mu \ne 4.55$

The average person watched 4.55 hours of television per day in 2005.

Part (a) Step 3. Compute the value of the test statistic.

Consider the MINITAB output, the value of test statistics is 0.41and the P-value is 0.687.

Using the rejection rule,

If $P\le \alpha$, then reject the null hypothesis.

It can be concluded that the P-value is 0 is greater than the level of significance, that is $P\left(=0.687\right)>\alpha \left(=0.10\right)$.

Therefore, the null hypothesis is not rejected at 10% level.

Hence, the data provide sufficient evidence to conclude that the amount of television watched per day last day year by the average person differed from that in2005.

Thus, both conclusions are same. It means, that the conclusion for confidence interval is same as the conclusion for hypothesis test.

Part (b) Step 1. Compute the confidence interval.

On computing the confidence interval,

1. Choose Stat>Basic Statistics>1-Sample t.
2. In Summarized data, enter the sample size 25 and mean 1922.76.
3. In Standard deviation, enter the value 350.90.
4. In Perform hypothesis test, enter the test mean as 1736.
5. Check Options, enter Confidence level as 95.
6. Choose not equal is alternative.
7. Click OK in all dialogue boxes.

Hence, from the MINITAB output, the 90% confidence interval is$\left(1777.9,2067.6\right)$.

Part (b) Step 2. Write the conclusion.

The population mean does not lies between lower and upper limit. Therefore, the null hypothesis is rejected at 5% level.

It can be concluded that the data provide sufficient evidence to conclude that the 2012 mean annual expenditure on apparel and services for consumer units in the Northeast differed from the national mean of $1736.

Using the Hypothesis test,

The null hypothesis is given below,

$H_0:\mu =1736$

The 2012 mean annual expenditure on apparel and services for consumer units in the Northeast does not differs from the national mean of $1736.

The alternative hypothesis is given below,

$H_a:\mu \ne \$1736$

The 2012 mean annual expenditure on apparel and services for consumer units in the Northeast is differs from the national mean of $1736.

Part (b) Step 3. Compute the value of the test statistic.

Consider the MINITAB output, the value of test statistics is 2.66and the P-value is 0.014.

Using the rejection rule,

If $P\le \alpha$, then reject the null hypothesis.

It can be concluded that the P-value is 0.014 is less than the level of significance, that is $P\left(=0.0137\right)<\alpha \left(=0.05\right)$.

Therefore, the null hypothesis is rejected at 5% level.

Hence, the data provide sufficient evidence to conclude that the 2012 mean annual expenditure on apparel and services for consumer units in the Northeast differed from the national mean of $1736.

Thus, both conclusions are same. It means, that the conclusion for confidence interval is same as the conclusion for hypothesis test.