To address this problem, we'll walk through the steps required to perform a hypothesis test for the variance:

Given:

• Previous variance ($\sigma_0^2$) = 53
• Sample variance ($s^2$) = 35
• Sample size ($n$) = 30
• Significance level ($\alpha$) = 0.05
• The problem involves testing whether the new variance is less than the previous variance, so it is a one-tailed test.

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(a) State the null and alternative hypotheses:

We are testing whether the variance has decreased.

• Null Hypothesis (H0): The variance is still 53, or $\sigma^2 = 53$.

  $H_0: \sigma^2 \geq 53$

• Alternative Hypothesis (H1): The variance has decreased, or $\sigma^2 < 53$.

  $H_1: \sigma^2 < 53$

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(b) Determine the type of test statistic to use:

Since we are testing the variance of a normally distributed population, we use a Chi-square test.

• Test statistic: $\chi^2 = \frac{(n-1) s^2}{\sigma_0^2}$

• Degrees of freedom (df): $n - 1 = 30 - 1 = 29$

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(c) Calculate the value of the test statistic:

$\chi^2 = \frac{(n-1) s^2}{\sigma_0^2} = \frac{(30-1) \cdot 35}{53} = \frac{29 \cdot 35}{53} = \frac{1015}{53} = 19.151$

The value of the test statistic is $\chi^2 = 19.151$.

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(d) Find the p-value:

Using a Chi-square distribution table or calculator, we look for the p-value associated with $\chi^2 = 19.151$ with 29 degrees of freedom for a one-tailed test (left-tailed since we're testing if the variance has decreased).

From the Chi-square distribution table:

• The p-value is approximately 0.137.

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(e) Conclusion:

Since the p-value (0.137) is greater than the significance level ($\alpha = 0.05$), we fail to reject the null hypothesis.

• Conclusion: There is not enough evidence to support the claim that the variance of customer ratings for the newly-priced deliveries is less than 53. Therefore, we cannot support the owner's claim.

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Summary:

• $H_0: \sigma^2 \geq 53$
• $H_1: \sigma^2 < 53$
• Test statistic $\chi^2 = 19.151$
• Degrees of freedom: 29
• p-value: 0.137
• Conclusion: No, we cannot support the claim.

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Do you need further explanation on any step? Here are 5 additional questions to explore this topic:

1. How do we determine the critical value for a Chi-square test?
2. What is the relationship between variance and standard deviation?
3. How would the conclusion change if the p-value was smaller than the significance level?
4. Can you describe why we used a Chi-square test and not a t-test in this case?
5. What are the assumptions underlying a Chi-square test for variance?

Tip: When calculating test statistics for variance, always ensure that the population is approximately normally distributed for the Chi-square test to be valid.