2.4 Hypothesis testing of single mean and single proportion: examples  (Page 3/6)

Set up the Hypothesis Test:

A 5% level of significance means that $\alpha =0.05$ . This is a test of a single population mean .

$H_{\mathrm{o}}$ : $\mu$ $=65$ $H_a$ : $\mu$ $>65$

Since the instructor thinks the average score is higher, use a " $>$ ". The " $>$ " means the test is right-tailed.

Determine the distribution needed:

Random variable: $\overline{X}$ = average score on the first statistics test.

Distribution for the test: If you read the problem carefully, you will notice that there is no population standard deviation given . You are only given $n=10$ sample data values. Notice also that the data come from a normal distribution. This means that thedistribution for the test is a student's-t.

Use $t_{\text{df}}$ . Therefore, the distribution for the test is $t_9$ where $n=10$ and $\text{df}=10-1=9$ .

Calculate the p-value using the Student's-t distribution:

$\text{p-value}=P($ $\overline{x}$ $>67$ ) $=0.0396$ where the sample mean and sample standard deviation are calculated as 67 and 3.1972 from the data.

Interpretation of the p-value: If the null hypothesis is true, then there is a 0.0396 probability (3.96%) that the sample mean is 67 or more.

Compare $\alpha$ and the p-value:

Since $\alpha =.05$ and $\text{p-value}=0.0396$ . Therefore, $\alpha >\text{p-value}$ .

Make a decision: Since $\alpha >\text{p-value}$ , reject $H_{\mathrm{o}}$ .

This means you reject $\mu =65$ . In other words, you believe the average test score is more than 65.

Conclusion: At a 5% level of significance, the sample data show sufficient evidence that the mean (average) test score is more than 65, just as the math instructor thinks.

The p-value can easily be calculated using the TI-83+ and the TI-84 calculators:

Put the data into a list. Press STAT and arrow over to TESTS . Press 2:T-Test . Arrow over to Data and press ENTER . Arrow down and enter 65 for ${\mu }_0$ , the name of the list where you put the data, and 1 for Freq: . Arrow down to $\mu :$ and arrow over to $>{\mu }_0$ . Press ENTER . Arrow down to Calculate and press ENTER . The calculator not only calculates the p-value ( $p=0.0396$ ) but it also calculates the test statistic (t-score) for the sample mean, the sample mean, and the sample standarddeviation. $\mu >65$ is the alternate hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate ). Press ENTER . A shaded graph appears with $t=1.9781$ (test statistic) and $p=0.0396$ (p-value). Make sure when you use Draw that no other equations are highlighted in $Y=$ and the plots are turned off.

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: $P\text{'}$ = 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 $P\text{'}$ , the estimated proportion.

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

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

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

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

Interpretation of the p-value: If the null hypothesis is true, there is 0.5485 probability (54.85%) that the sample (estimated) proportion $p\text{'}$ 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.

$p\text{'}$ $=0.53$ . 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 p-value can easily be calculated using the TI-83+ and the TI-84 calculators:

Press STAT and arrow over to TESTS . Press 5:1-PropZTest . Enter .5 for $p_0$ , 53 for $x$ and 100 for $n$ . Arrow down to Prop and arrow to not equals $p_0$ . Press ENTER . Arrow down to Calculate and press ENTER . The calculator calculates the p-value ( $p=0.5485$ ) and the test statistic (z-score). Prop not equals .5 is the alternate hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate ). Press ENTER . A shaded graph appears with $z=0.6$ (test statistic) and $p=0.5485$ (p-value). Make sure when you use Draw that no other equations are highlighted in $Y=$ and the plots are turned off.

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.)