9.11 Hypothesis testing of single mean and single proportion:  (Page 2/4)

Set up the Hypothesis Test:

The 1% level of significance means that $\alpha =0.01$ . This is a test of a single population proportion .

$H_{\mathrm{o}}$ : $p$ $=0.50$ $H_a$ : $p$ $\ne 0.50$

The words "is the same or different from" tell you this is a two-tailed test.

Calculate the distribution needed:

Random variable: $\hat{p}$ = the percent of of first-time brides who are younger than their grooms.

Distribution for the test: The problem contains no mention of a mean. The information is given in terms of percentages. Use the distribution for $\hat{p}$ , the estimated proportion.

$\hat{p}$ ~ $N$ $(p,\sqrt{\frac{p\cdot q}{n}})$ Therefore, $\hat{p}$ ~ $N$ $(0.5,\sqrt{\frac{0.5\cdot 0.5}{100}})$ where $p=0.50$ , $q=1-p=0.50$ , and $n=100$ .

Calculate the p-value using the normal distribution for proportions:

$\text{p-value}=P(\hat{p}$ $$ $0.47$ or $\hat{p}>0.53$ ) $=0.5485$

where $x=53$ , $\hat{p}=\frac{x}{n}$ $=\frac{53}{100}=0.53$ .

Interpretation of the p-value: If the null hypothesis is true, there is 0.5485 probability (54.85%) that the sample (estimated) proportion $\hat{p}$ is 0.53 or more OR 0.47 or less (see the graph below).

$\mu =p=0.50$ comes from $H_{\mathrm{o}}$ , the null hypothesis.

$\hat{p}$ $=0.53$ . Since the curve is symmetrical andthe test is two-tailed, the $p\text{'}$ for the left tail is equal to $0.50-0.03=0.47$ where $\mu =p=0.50$ . (0.03 is the differencebetween 0.53 and 0.50.)

Compare $\alpha$ and the p-value:

Since $\alpha =0.01$ and $\text{p-value}=0.5485$ . Therefore, $\alpha$ $$ $\text{p-value}$ .

Make a decision: Since $\alpha$ $$ $\text{p-value}$ , you cannot reject $H_{\mathrm{o}}$ .

Conclusion: At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of first-time brides that are younger than their grooms isdifferent from 50%.

The Type I and Type II errors are as follows:

The Type I error is to conclude that the proportion of first-time brides that are younger than their grooms is different from 50% when, in fact, the proportion is actually 50%.(Reject the null hypothesis when the null hypothesis is true).

The Type II error is there is not enough evidence to conclude that the proportion of first time brides that are younger than their grooms differs from 50% when, in fact, the proportion does differ from 50%. (Do not reject the null hypothesis when the null hypothesis is false.)

Set up the Hypothesis Test:

$H_{\mathrm{o}}$ : $p=0.25$ $H_{\mathrm{a}}$ : $p>0.25$

Determine the distribution needed:

In words, CLEARLY state what your random variable $\overline{X}$ or $\hat{p}$ represents.

$\hat{p}$ = The proportion of fleas that are killed by the new shampoo

State the distribution to use for the test.

Normal: $N(0.25,\sqrt{\frac{\left(0.25\right)(1-0.25)}{42}})$

Test Statistic: $z=2.3163$

Calculate the p-value using the normal distribution for proportions:

$\text{p-value}=$ $0.0103$

In 1 – 2 complete sentences, explain what the p-value means for this problem.

If the null hypothesis is true (the proportion is 0.25), then there is a 0.0103 probability that the sample (estimated) proportion is 0.4048 $\left(\frac{17}{42}\right)$ or more.

Use the previous information to sketch a picture of this situation. CLEARLY, label and scale the horizontal axis and shade the region(s) corresponding to the p-value.

Compare $\alpha$ and the p-value:

Indicate the correct decision (“reject” or “do not reject” the null hypothesis), the reason for it, and write an appropriate conclusion, using COMPLETE SENTENCES.

Conclusion: At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of fleas that are killed by the new shampoo is more than 25%.

Construct a 95% Confidence Interval for the true mean or proportion. Include a sketch of the graph of the situation. Label the point estimate and the lower and upper bounds of the Confidence Interval.

Confidence Interval: $(0.26,0.55)$ We are 95% confident that the true population proportion $p$ of fleas that are killed by the new shampoo is between 26% and 55%.