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A researcher takes sample temperatures in Fahrenheit of 20 days from Cleveland and 19 days from Memphis. Test the claim that the mean temperature in Cleveland greater than the mean temperature in Memphis. Use a significance level of 
. Assume the populations are approximately normally distributed with unequal variances. You obtain the following two samples of data. Round answers to 4 decimal places.

Cleveland	Memphis
77	68.1
93	66.9
60.7	68.5
52.9	73.4
72.6	76
64.9	64.2
110.5	65.7
40.6	81.7
107.2	74.9
57.1	48.5
101.1	55.1
67.9	63.5
80	74.4
83.3	41.3
89	49.6
58.4	53.9
56.5	64.2
63.4	80.8
62.9	60.9
78.5	
What are the correct hypotheses? Note this may view better in full screen mode. Select the correct symbols for each of the 6 spaces.

H0: 
Select an answer
 
?
 
Select an answer


Ha: 
Select an answer
 
?
 
Select an answer


Based on the hypotheses, find the following:

Test Statistic = 


p-value = 

The p-value is 
Select an answer
 to alpha.

The correct decision is to 
Select an answer
 .

The correct summary would be: 
Select an answer
 the claim that the mean temperature in Cleveland greater than the mean temperature in Memphis.

To test the claim that the mean temperature in Cleveland is greater than the mean temperature in Memphis, we perform a one-tailed two-sample t-test assuming unequal variances. Let's go through the steps to solve this problem:

### Step 1: State the Hypotheses
The null hypothesis (H₀) and the alternative hypothesis (Ha) are as follows:
- \( H_0 \): \( \mu_{\text{Cleveland}} \leq \mu_{\text{Memphis}} \)
- \( H_a \): \( \mu_{\text{Cleveland}} > \mu_{\text{Memphis}} \)

### Step 2: Calculate the Sample Means and Standard Deviations
First, let's calculate the sample means and standard deviations for the temperatures in Cleveland and Memphis.

**Cleveland:**
\[
\text{Mean of Cleveland (}\overline{X_1}\text{)} = \frac{\sum X_1}{n_1} = \frac{77 + 93 + 60.7 + 52.9 + 72.6 + 64.9 + 110.5 + 40.6 + 107.2 + 57.1 + 101.1 + 67.9 + 80 + 83.3 + 89 + 58.4 + 56.5 + 63.4 + 62.9 + 78.5}{20} 
\]
\[
= \frac{1398.4}{20} = 69.92
\]

\[
\text{Standard Deviation of Cleveland (} s_1 \text{)} = \sqrt{\frac{\sum (X_1 - \overline{X_1})^2}{n_1-1}}
\]

**Memphis:**
\[
\text{Mean of Memphis (}\overline{X_2}\text{)} = \frac{\sum X_2}{n_2} = \frac{68.1 + 66.9 + 68.5 + 73.4 + 76 + 64.2 + 65.7 + 81.7 + 74.9 + 48.5 + 55.1 + 63.5 + 74.4 + 41.3 + 49.6 + 53.9 + 64.2 + 80.8 + 60.9}{19} 
\]
\[
= \frac{1231.5}{19} = 64.8158
\]

\[
\text{Standard Deviation of Memphis (} s_2 \text{)} = \sqrt{\frac{\sum (X_2 - \overline{X_2})^2}{n_2-1}}
\]

### Step 3: Calculate the Test Statistic
The formula for the test statistic when assuming unequal variances (Welch's t-test) is:

\[
t = \frac{\overline{X_1} - \overline{X_2}}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}
\]

To calculate the standard deviations \( s_1 \) and \( s_2 \):

\[
s_1^2 = \frac{1}{n_1-1} \sum (X_1 - \overline{X_1})^2
\]
\[
s_2^2 = \frac{1}{n_2-1} \sum (X_2 - \overline{X_2})^2
\]

Using the sample data, the standard deviations are:
\[
s_1 \approx 21.49, \quad s_2 \approx 10.56
\]

Now, substitute these values into the test statistic formula:

\[
t = \frac{69.92 - 64.8158}{\sqrt{\frac{21.49^2}{20} + \frac{10.56^2}{19}}}
\]
\[
t = \frac{5.1042}{\sqrt{\frac{461.2801}{20} + \frac{111.5136}{19}}}
\]
\[
t = \frac{5.1042}{\sqrt{23.064 + 5.8681}}
\]
\[
t = \frac{5.1042}{\sqrt{28.9321}}
\]
\[
t = \frac{5.1042}{5.3797}
\]
\[
t \approx 0.9486
\]

### Step 4: Determine the Degrees of Freedom
Using the formula for degrees of freedom in Welch's t-test:

\[
df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1-1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2-1}}
\]

\[
df = \frac{\left(\frac{21.49^2}{20} + \frac{10.56^2}{19}\right)^2}{\frac{\left(\frac{21.49^2}{20}\right)^2}{19} + \frac{\left(\frac{10.56^2}{19}\right)^2}{18}}
\]
\[
df \approx 28.9
\]

### Step 5: Find the p-value
Using a t-distribution table or calculator for \( t = 0.9486 \) and \( df \approx 28.9 \):

\[
p \text{-value} \approx 0.176
\]

### Step 6: Make the Decision
Since the p-value (0.176) is greater than the significance level (0.10), we fail to reject the null hypothesis.

### Conclusion
There is not enough evidence to support the claim that the mean temperature in Cleveland is greater than the mean temperature in Memphis.

---

**Test Statistic:** 0.9486

**p-value:** 0.176

**The p-value is greater than alpha.**

**Decision:** Do not reject the null hypothesis.

**Summary:** There is not enough evidence to support the claim that the mean temperature in Cleveland is greater than the mean temperature in Memphis.

---

Would you like any further details or have any questions about this process?

Here are 8 related questions to explore further:
1. How do you calculate the sample standard deviation?
2. What assumptions are necessary for conducting a t-test?
3. How does a one-tailed test differ from a two-tailed test in terms of hypothesis and p-value?
4. What are the potential consequences of incorrectly rejecting the null hypothesis?
5. How does the significance level affect the conclusion of a hypothesis test?
6. What is the importance of sample size in hypothesis testing?
7. How can you interpret the results of a t-test in real-world scenarios?
8. What are some common pitfalls to avoid when performing hypothesis testing?

**Tip:** Always ensure your data meets the assumptions of the statistical test you are using to avoid incorrect conclusions.

Explore a hypothesis test comparing mean temperatures between Cleveland and Memphis using a two-sample t-test. Calculate the test statistic and p-value to determine if there is significant evidence that Cleveland's mean temperature is greater. Includes step-by-step calculations and analysis.

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