Question

11.93 Washing Up. A Harris Interactive survey found that $92.0\%$ of $1001$American adults said they always wash up after using the bathroom.
a. At the $5\%$significance level, do the data provide sufficient evidence to conclude that more than $9$of $10$Americans always wash up after using the bathroom?
b. Repeat part (a), using a $1\%$level of significance.

Short answer

(a) At the $5\%$level, the test results are statistically significant. Yes, the data supports the conclusion that more than $9$out of $10$Americans are always awake after using the bathroom.

(b) At the $1\%$level, the test result is not statistically significant. The data does not support the conclusion that more than $9$out of $10$Americans are always awake after using the bathroom.

Step-by-step solution

Part (a) Step 1: Given information

To find that the data provide sufficient evidence to conclude that more than $9$of $10$Americans always wash up after using the bathroom at the significance level of $5\%$.

Part (a) Step 2: Explanation

Since, the significance level is $5\%,n=1001,p_0=0.9,$and $\hat{p}=0.92$
Determine the value of $n{p}_0$ as:
$n{p}_0=\left(1001\right)(0.9)$
$=900.9$
Determine the value of $n\left(1-p_0\right)$ as:
$n\left(1-p_0\right)=\left(1001\right)(1-0.9)$
$=100.1$
The values of $n{p}_0$and $n(1-p_0)$ are both larger than $5$.
As a result, only one proportion $z$ test should be used.
The null hypothesis: $H_0:p_0=0.9$
The alternate hypothesis: $H_{\mathrm{a}}:p_0>0.9$

Part (a) Step 3: Explanation

Determine the $z$value as:

$z=\frac{\hat{p}-p_0}{\sqrt{\frac{P_0\left(1-P_0\right)}{n}}}$
$=\frac{0.920-0.9}{\sqrt{{\displaystyle \frac{0.9(1-0.9)}{1001}}}}$
$=\frac{0.020}{0.009}$
$=2.11$
Since,$\alpha =0.05$
The test is right tailed.
The critical value of $z$, for $\alpha =0.05$, from the standard table as:
$z_{0.05}=1.645$
The test statistic falls in the rejection region.

As a result, the hypothesis $H_0$is rejected.
At the$5\%$ level, the test results are statistically significant.
As a result, yes, the data supports the conclusion that more than $9$ out of $10$ Americans are always awake after using the bathroom.

Part (b) Step 1: Given information

To find that the data provide sufficient evidence to conclude that more than $9$of $10$ Americans always wash up after using the bathroom using the significance level of $1\%$.

Part (b) Step 2: Explanation

Since, the significance level is $1\%,n=1001,p_0=0.9,$and $\hat{p}=0.92$.

Determine the value of $n{p}_0$as:

$n{p}_0=\left(1001\right)(0.9)$

$=900.9$

Determine the value of $n(1-p_0)$as:

$n(1-p_0)=\left(1000\right)(1-0.9)$

$=100.1$

The values of $n{p}_0$and $n(1-p_0)$ are both larger than $5$.
As a result, only one proportion $z$ test should be used.
The null hypothesis: $H_0:p_0=0.9$
The alternate hypothesis: $H_{\mathrm{a}}:p_0>0.9$

Part (b) Step 3: Explanation

Determine the value for $z$ as:
$z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}$
$z=\frac{0.920-0.9}{\sqrt{{\displaystyle \frac{0.9(1-0.9)}{1001}}}}$
$=\frac{0.020}{0.009}$
$=2.11$

Since,$\alpha =0.01$
The test is right tailed.
From the standard table, the critical value of $z$for $\alpha =0.01$as:
$z_{0.01}=2.33$
The test statistic falls in the acceptance region.

Asa result, the hypothesis $H_0$ is not rejected.
At the$1\%$ level, the test result is not statistically significant.
As a result, the data does not support the conclusion that more than $9$ out of $10$Americans are always awake after using the bathroom.