Posted by **Sheila** on Friday, April 20, 2012 at 7:20pm.

An automobile dealership has as one of its performance goals that the proportion of its automobiles sold that are deemed “good value for the money” by the purchasers be at least .90. For a random sample of 400 automobiles sold over the past six months, 352 were deemed by the purchasers to be “good value for the money.” Does this sample data provide evidence that the dealership is not meeting its performance goal? Either use the p-value approach to hypothesis testing or use the significance level approach with á = .05.

- Statistics -
**MathGuru**, Friday, April 20, 2012 at 7:38pm
Try a proportional one-sample z-test for this one since this problem is using proportions.

Here's a few hints to get you started:

Null hypothesis:

Ho: p ≥ .90 -->meaning: population proportion is greater than or equal to .90

Alternate hypothesis:

Ha: p < .90 -->meaning: population proportion is less than .90

Using a formula for a proportional one-sample z-test with your data included:

z = .88 - .90 -->test value (352/400 = .88) minus population value (.90) divided by

√[(.90)(.10)/400] --> .10 represents 1-.90 and 400 is sample size.

Finish the calculation. Remember if the null is not rejected, then there is no difference. If you need to find the p-value for the test statistic, check a z-table. The p-value is the actual level of the test statistic. Otherwise, check a z-table for a one-tailed test at .05 level of significance. Compare to your test statistic. If the test statistic exceeds the value from the table, reject the null and conclude p < .90. If the test statistic does not exceed the value from the table, do not reject the null.

I hope this will help.

- Statistics -
**Sheila**, Friday, April 20, 2012 at 8:33pm
Thanks MathGuru! Makes a ton more sense now!

- Statistics -
**Srinivas**, Monday, August 13, 2012 at 7:49pm
MathGuru,

/****

statistic. Otherwise, check a z-table for a one-tailed test at .05 level of significance. Compare to your test statistic. If the test statistic exceeds the value from the table, reject the null and conclude p < .90. If the test statistic does not exceed the value from the table, do not reject the null.

****/

I think you got the opposite.

Decision rule should be--> Accept Ho if test statistic >= -1.645

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