Question

Expand Suppose that heights of male collegiate basketball players in a country are normally distributed with a mean of 75 in and a standard deviation of 3.6 in. A researcher wants to determine if the mean height of male collegiate basketball players in one particular conference is different from the national average. She obtains email addresses for all of the players in this conference and emails them asking for them to reply with their height Of the 573 emails she sent, she obtained 24 replies. Because she does not have the data for all 573, she decides to use a one- sample z-test for a population mean using the standard deviation of all male collegiate basketball players in the country The heights, in inches, for her sample are listed below. 78, 73, 79, 82,81,75, 73, 83, 75, 77,79, 80, 78, 78, 77, 79, 78, 74, 79, 75, 73, 80, 79, 78 Is the researcher justified in using the one-sample z-test for a population mean? O No, because the sample is based, and is therefore not a simple random sample. O Yes, all requirements are met. The population standard deviation is known, the population is approximately normal, the sample is large enough, random, and contains no outliers O No, because the sample data has a significant outlier, indicating that the sample may not come from a normal distribution. O No, because the population standard deviation is not known. O No, because the population of all male collegiate basketball players in the conference is not known to be normally distributed. about us careers privacy policy terms of use contact us help

Answer 1

One sample Z-test is appropriate test for this condition.

Yes, all the requirements are met. The population standard deviation is known, the population is approximately normal, the sample is large enough, random, and contains no outliers.

Wesolve this problem by using MINITAB

Here population standard deviation ( $\sigma$ ) is known, so we can used one sample z test

Let's write the given information.

level of significance = $\alpha$ = 5%

Let's use minitab

The command is Stat>>>Basic Statistics >>1 sample Z...

Click on Sample in Columns

Select data set

standard deviation = 3.6

then click oon "Perform hypothesis test

Hypothesized mean = 75 (from null hypothesis)

then click on Option

Confidence level = 100 - $\alpha$ % =100 - 5 = 95.0

Alternative " not equal"

then click on OK

again click on OK

We get the following output

One-Sample Z: Heights

Test of mu = 75 vs not = 75
The assumed standard deviation = 3.6

Variable   N   Mean StDev    SE Mean       95% CI          Z         P
Heights   24 77.625 2.795    0.735    (76.185, 79.065) 3.57 0.000

Decision rule: 1) If p-value $\leq$ level of significance ( $\alpha$ ) then we reject null hypothesis

2) If p-value > level of significance (alpha) then we fail to reject null hypothesis.

Here p value = 0.000 < 0.05 so we used first rule.

That is we reject null hypothesis

Conclusion: At 5% level of significance there are sufficient evidence to say that the mean height of male collegiate basketball players in one perticular conference is different from the national average.