This problem requires performing a hypothesis test to determine if the variance in customer satisfaction ratings has decreased after the price change. We will conduct a one-tailed test for variance. Here's how we approach the problem step by step:

(a) State the null hypothesis $H_0$ and the alternative hypothesis $H_1$.

The hypotheses are based on the claim about the variance $\sigma^2$ of customer satisfaction ratings:

• $H_0$: $\sigma^2 = 53$ (The variance is equal to the original variance)
• $H_1$: $\sigma^2 < 53$ (The variance has decreased)

This is a one-tailed test because we are only interested in whether the variance has decreased, not if it has simply changed.

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

Since we are testing a hypothesis about a population variance, we will use the chi-square test statistic. The formula for the chi-square statistic when testing variance is:

$\chi^2 = \frac{(n - 1) \cdot s^2}{\sigma_0^2}$

Where:

• $n = 30$ is the sample size,
• $s^2 = 35$ is the sample variance,
• $\sigma_0^2 = 53$ is the variance under the null hypothesis.

(c) Find the value of the test statistic.

We can now plug in the values into the chi-square formula:

$\chi^2 = \frac{(30 - 1) \cdot 35}{53} = \frac{29 \cdot 35}{53}$

Calculating this gives:

$\chi^2 = \frac{1015}{53} = 19.1509$

So, the value of the test statistic is $\chi^2 \approx 19.151$.

(d) Find the p-value.

The p-value is the probability of observing a test statistic as extreme as (or more extreme than) the one calculated, assuming the null hypothesis is true. This is a one-tailed test with 29 degrees of freedom (since $n-1 = 30-1$).

Using a chi-square distribution table or a statistical calculator, we find the p-value for $\chi^2 = 19.151$ with 29 degrees of freedom. The p-value is approximately 0.103.

(e) Can we support the claim that the variance of customer ratings for the newly-priced deliveries is less than 53?

The significance level $\alpha = 0.05$. Since the p-value (0.103) is greater than $\alpha$, we fail to reject the null hypothesis.

Thus, there is not enough evidence to support the owner's claim that the variance has decreased.

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Do you want more details on any of the steps or calculations?

Related Questions:

1. How does the chi-square distribution change as the degrees of freedom increase?
2. What are the assumptions required for a chi-square test of variance?
3. How is the critical value for a chi-square test determined?
4. What is the difference between a one-tailed and two-tailed test in hypothesis testing?
5. Can sample variance ever be negative? Why or why not?

Tip:

When conducting hypothesis tests involving variance, always ensure that the data is approximately normally distributed, as this is an important assumption for the chi-square test.