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.
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.
Thanks MathGuru! Makes a ton more sense now!
MathGuru,
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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.
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I think you got the opposite.
Decision rule should be--> Accept Ho if test statistic >= -1.645
To determine if the dealership is meeting its performance goal, we can conduct a hypothesis test using the p-value approach.
Step 1: State the hypotheses.
- Null hypothesis: The proportion of automobiles deemed "good value for the money" is equal to or greater than 0.90. (H₀: p ≥ 0.90)
- Alternative hypothesis: The proportion of automobiles deemed "good value for the money" is less than 0.90. (H₁: p < 0.90)
Step 2: Set the significance level.
We are given α = 0.05, which means we need enough evidence to reject the null hypothesis with a 95% confidence level.
Step 3: Calculate the test statistic.
We'll use a one-sample proportion test. The test statistic formula is:
z = (p̂ - p₀) / √(p₀(1-p₀) / n)
where p̂ is the sample proportion, p₀ is the hypothesized proportion, and n is the sample size.
In this case, p̂ = 352/400 = 0.88 (proportion of automobiles deemed "good value for the money" in the sample), p₀ = 0.90 (hypothesized proportion), and n = 400.
Plugging these values into the formula, we get:
z = (0.88 - 0.90) / √(0.90 * (1 - 0.90) / 400) ≈ -0.632
Step 4: Calculate the p-value.
The p-value is the probability of observing a test statistic as extreme as the one calculated or more extreme, assuming the null hypothesis is true. In this case, we want to find P(Z < -0.632).
Using a standard normal distribution table or a calculator, we find that P(Z < -0.632) ≈ 0.2648.
Step 5: Draw a conclusion.
Comparing the p-value (0.2648) to the significance level (α = 0.05), we see that the p-value is greater than α. Therefore, we fail to reject the null hypothesis.
Conclusion:
There is not enough evidence to conclude that the dealership is not meeting its performance goal. The sample data does not provide sufficient evidence to support the claim that the proportion of automobiles deemed "good value for the money" is less than 0.90.