Chapter 10.4, Problem 7E

To determine

To perform each hypothesis test and to complete the following steps:

(a). To state the null and alternative hypotheses.

(b). To determine which distribution to use for the test statistic and to state the level of significance.

(c). To calculate the test statistic.

(d). To draw a conclusion and by comparing the $p$-value to the level of significance to interpret the decision.

Answer to Problem 7E

Solution:

(a). The null and alternative hypotheses.

$H_o:p=0.$ 058

$H_a:p\ne 0.058$

(b). The distribution to use for the test statistic and the level of significance.

Normal distribution and the $z-$ test static for the sample proportion.

Level of significance $\alpha =0.01$

(c). The test statistic.

$z\approx 2.22$

(d). Conclusion and by comparing the $p$-value to the level of significance the interpretation of the decision.

We fail to reject the null hypothesis.

This means that at the $0.01$ level of significance, the evidence does not supports safety commission’s claim.

Explanation of Solution

Approach:

The null hypotheses is $H_o:p$ related by greater than or lesser than and will contain equality sign but alternate hypotheses $H_a:p$ is just a logical opposite of null hypotheses.

Conditions to meet for using normal distribution to perform a hypothesis test for the population proportion:

All possible samples of a given size have an equal probability of being chosen; that is, a simple random sample is used.

The conditions for a binomial distribution are met.

The sample size is large enough to ensure that $np\ge 5$ and $n\left(1-p\right)\ge 5$.

Level of significance

$\alpha =1-c$

Sample proportion.

$\widehat{p}=\frac{Total out come}{Sample size}$

The test statistic is given by the $z-$ score.

$z=\frac{\widehat{p}-p}{\sqrt{\frac{p\left(1-p\right)}{n}}}$

Decision.

$p-$ value $>\alpha$	Fail to reject the null hypothesis.
$p-$ value $\le \alpha$	Reject the null hypothesis.

Calculation:

Given,

Type of sample: simple random sample

Number of drivers surveyed $=350$

Number of drivers suffer from sleep apnea $= 30$

Percentage of Americans suffer from sleep apnea is 5.8%.

Safety commission’s claim: Percentage of Americans suffer from sleep apnea is not 5.8%

Level of significance $=0.01$

Claim	Population proportion
Number of trails (sample size)	$n=350$
Probability of success that the claim referencing	$p=\frac{5.8}{100}=0.058$
Level of significance	$\alpha =0.01$

Step (a):

Null and alterative hypothesis:

The safety commission’s claim is that the percentage of Americans suffer from sleep apnea is not 5.8%

Mathematically we can write $p\ne 0.058$

The logical opposite of this claim is $p=0.058$

Thus the null and alterative hypothesis are stated as follows.

$H_o:p=0.$ 058

$H_a:p\ne 0.058$

Step (b):

Distribution to use for the test statistic and level of significance:

We are testing a population proportion, se we must check the necessary conditions to use the normal distribution and the $z-$ test static.

Test for normal distribution:

Let us check that whether the sample size is large enough to ensure that $np \ge 5$ and $n\left(1-p\right)\ge 5$.

Calculate: $np$

$= np$

$=350 \left(0.058\right)$

$=20.3>5$

Thus, $np \ge 5$ is satisfied.

Calculate: $n(1-p)$

$=n(1-p)$

$=350 (1-0.058)$

$=350 \left(0.942\right)$

$=329.7>5$

Also $n(1-p)\ge 5$ is satisfied.

Since, all the conditions are satisfied, we can use the normal distribution and the $z-$ test static for the sample proportion.

Level of significance $\alpha =0.01$

Step (c):

Test static $\left(z\right)$ calculation:

First let us compute the value of sample proportion.

In the sample data $30$ drivers out of $350$ were suffered by sleep apnea.

Thus,

$\widehat{p}=\frac{30}{350}$

$\widehat{p}=0.085714$

Now; substitute the values

$\widehat{p}=0.085714,$

$p=0.058,$ and

$n=350$ in the following $\mathrm{z}-$ score to get

$z=\frac{\widehat{p}-p}{\sqrt{\frac{p\left(1-p\right)}{n}}}$

$z=\frac{0.085714-0.058}{\sqrt{\frac{0.058\left(1-0.058\right)}{350}}}$

$z=\frac{0.027714}{\sqrt{\frac{0.058\left(0.942\right)}{350}}}$

$z=\frac{0.027714}{\sqrt{\frac{0.054636}{350}}}$

$z=\frac{0.027714}{\sqrt{0.0001561}}$

$z=\frac{0.027714}{0.012494}$

$z=2.218184$

$z\approx 2.22$

Step (d):

Draw a conclusion and to interpret the decision by comparing the p-value to the level of significance:

The alternative hypothesis tells us that we are conducting a two tailed test.

Thus, the p- value for this test statistic is the probability of obtaining a test statistic is either less than or equal to $z_1=-2.22$ or greater than or equal to $z_2=2.22$ written mathematically p-value $=P(\left|z\right|\ge 2.22)$

To find the p-value, we need to find the area under the standard normal curve to the left of $z_1=-2.22$

Hence, the area under the normal distribution curve the normal distribution curve to the left of $-2.22$ is given by

$P\left(z_1\le -2.22\right)=0.0264$.

Thus the total area is the twice the area of the left tail.

p-value $=P\left(\left|z\right|\ge 2.22\right)=2\left(0.0264\right)$

$p=0.0528$

Compare the p-value to the level significance.

$0.0528>0.01$

(i.e.) $p-$ value $>\alpha$

Thus, we fail to reject the null hypothesis.

This means that at the $0.01$ level of significance, the evidence does not supports safety commission’s claim.

Final statement:

Therefore,

(a). The null and alternative hypotheses.

$H_o:p=0.058$

$H_a:p\ne 0.058$

(b). The distribution to use for the test statistic and the level of significance.

Normal distribution and the $z-$ test static for the sample proportion.

Level of significance $\alpha =0.01$

(c). The test statistic.

$z\approx 2.22$

(d). Conclusion and by comparing the $p$-value to the level of significance the interpretation of the decision.

We fail to reject the null hypothesis.

This means that at the $0.01$ level of significance, the evidence does not supports safety commission’s claim.