"Testing the Difference Between Two Means In Exercises 15–24, (a) identify the claim and state Ho and Ha, (b) find the critical value(s) and identify the rejection region(s), (c) find the standardized test statistic z, (d) decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim. Assume the samples are random and independent, and the populations are normally distributed.
[APPLET] Precipitation A climatologist claims that the precipitation in Seattle, Washington, was greater than in Birmingham, Alabama, in a recent year. The daily precipitation amounts (in inches) for 30 days in a recent year in Seattle are shown below. Assume the population standard deviation is 0.25 inch.

0.00 0.00 0.05 0.01 0.21 0.00 0.00 0.52 0.00 0.010.00 0.19 0.00 0.18 0.02 0.02 0.13 0.00 0.03 0.000.04 0.00 0.41 0.23 0.00 0.80 0.15 0.00 0.00 0.79


The daily precipitation amounts (in inches) for 30 days in a recent year in Birmingham are shown below. Assume the population standard deviation is 0.52 inch.


0.00 0.96 0.84 0.00 0.10 0.00 0.00 0.20 0.00 0.54 0.97 0.00 0.35 0.02 0.04 0.70 0.00 0.00 0.00 0.00 0.03 0.01 0.15 0.27 0.00 0.00 0.93 0.00 0.89 0.01


At α=0.05, can you support the climatologist’s claim? (Source: NOAA)"

Verified step by step guidance

1. Identify the claim and state the null hypothesis (H\_0) and alternative hypothesis (H\_a). Since the climatologist claims that precipitation in Seattle is greater than in Birmingham, the hypotheses are: \(H_0: \mu_{Seattle} \leq \mu_{Birmingham}\) \(H_a: \mu_{Seattle} > \mu_{Birmingham}\)
2. Determine the significance level \( \alpha = 0.05 \) and find the critical value for a right-tailed z-test. Use the standard normal distribution table to find the z-value such that the area to the right is 0.05. This z-value defines the rejection region where you would reject the null hypothesis.
3. Calculate the sample means for Seattle and Birmingham from the given data. Then, compute the standardized test statistic \( z \) using the formula: \(z = \frac{\bar{x}_{Seattle} - \bar{x}_{Birmingham}}{\sqrt{\frac{\sigma_{Seattle}^2}{n_{Seattle}} + \frac{\sigma_{Birmingham}^2}{n_{Birmingham}}}}\) where \( \sigma_{Seattle} = 0.25 \), \( \sigma_{Birmingham} = 0.52 \), and both sample sizes \( n_{Seattle} = n_{Birmingham} = 30 \).
4. Compare the calculated z-value to the critical value found in step 2. If the z-value is greater than the critical value, reject the null hypothesis; otherwise, fail to reject it.
5. Interpret the decision in the context of the original claim. If you rejected the null hypothesis, you have sufficient evidence at the 0.05 significance level to support the climatologist's claim that Seattle's precipitation is greater than Birmingham's. If you failed to reject, there is not enough evidence to support the claim.

Key Concepts

Hypothesis Testing for Two Means

This involves comparing the means of two independent populations to determine if there is a statistically significant difference. The null hypothesis (H0) usually states that the means are equal, while the alternative hypothesis (Ha) reflects the claim, such as one mean being greater. This framework guides the testing process and decision-making.

Critical Value and Rejection Region

The critical value is a threshold determined by the significance level (α) that defines the boundary of the rejection region. If the test statistic falls into this region, the null hypothesis is rejected. For a one-tailed test at α=0.05, the critical z-value marks the cutoff for deciding if the observed data is unlikely under H0.

Standardized Test Statistic (z-score)

The z-score measures how many standard errors the sample mean difference is from the hypothesized difference (usually zero). It standardizes the difference using known population standard deviations and sample sizes, allowing comparison to the normal distribution to assess significance.