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Answer :
To solve this problem, we are testing the claim that the mean temperature in New Orleans is different from the mean temperature in Milwaukee, using a significance level of [tex]\(\alpha=0.10\)[/tex]. We will assume that the populations are approximately normally distributed with unequal variances. This means we can use a two-sample t-test for unequal variances, also known as Welch's t-test.
Here’s a step-by-step explanation of the solution:
1. State the Hypotheses:
- Null Hypothesis ([tex]\(H_0\)[/tex]): The mean temperature in New Orleans is equal to the mean temperature in Milwaukee.
[tex]\[
H_0: \mu_1 = \mu_2
\][/tex]
- Alternative Hypothesis ([tex]\(H_1\)[/tex]): The mean temperature in New Orleans is different from the mean temperature in Milwaukee.
[tex]\[
H_1: \mu_1 \neq \mu_2
\][/tex]
2. Calculate the Sample Means and Standard Deviations:
- Mean temperature for New Orleans: [tex]\( \bar{x}_1 = 63.5100 \)[/tex]
- Mean temperature for Milwaukee: [tex]\( \bar{x}_2 = 76.5480 \)[/tex]
- Standard deviation for New Orleans: [tex]\( s_1 = 13.9203 \)[/tex]
- Standard deviation for Milwaukee: [tex]\( s_2 = 18.7426 \)[/tex]
3. Determine the Sample Sizes:
- Number of days sampled in New Orleans, [tex]\( n_1 = 20 \)[/tex]
- Number of days sampled in Milwaukee, [tex]\( n_2 = 25 \)[/tex]
4. Calculate the Test Statistic:
The test statistic for Welch's t-test is calculated using the formula:
[tex]\[
t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}
\][/tex]
Substituting the given values:
[tex]\[
t = \frac{63.5100 - 76.5480}{\sqrt{\frac{13.9203^2}{20} + \frac{18.7426^2}{25}}} = -2.6759
\][/tex]
5. Calculate the Degrees of Freedom:
The degrees of freedom for Welch's t-test is given by:
[tex]\[
df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1-1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2-1}}
\][/tex]
The calculated degrees of freedom is approximately:
[tex]\[
df \approx 42.8024
\][/tex]
6. Find the p-value:
The p-value is calculated using the cumulative distribution function (CDF) for the t-distribution:
[tex]\[
p \text{-value} = 2 \times \left(1 - \text{CDF}(|t|, df)\right)
\][/tex]
With [tex]\(t = -2.6759\)[/tex] and [tex]\(df = 42.8024\)[/tex], the p-value is approximately 0.0105.
7. Make the Decision:
Compare the p-value to the significance level [tex]\(\alpha = 0.10\)[/tex]:
- Since [tex]\( \text{p-value} = 0.0105 < \alpha = 0.10 \)[/tex], we reject the null hypothesis.
8. Conclusion:
The conclusion is that there is enough evidence at the 0.10 significance level to reject the null hypothesis. Therefore, we conclude that the mean temperature in New Orleans is different from the mean temperature in Milwaukee.
Here’s a step-by-step explanation of the solution:
1. State the Hypotheses:
- Null Hypothesis ([tex]\(H_0\)[/tex]): The mean temperature in New Orleans is equal to the mean temperature in Milwaukee.
[tex]\[
H_0: \mu_1 = \mu_2
\][/tex]
- Alternative Hypothesis ([tex]\(H_1\)[/tex]): The mean temperature in New Orleans is different from the mean temperature in Milwaukee.
[tex]\[
H_1: \mu_1 \neq \mu_2
\][/tex]
2. Calculate the Sample Means and Standard Deviations:
- Mean temperature for New Orleans: [tex]\( \bar{x}_1 = 63.5100 \)[/tex]
- Mean temperature for Milwaukee: [tex]\( \bar{x}_2 = 76.5480 \)[/tex]
- Standard deviation for New Orleans: [tex]\( s_1 = 13.9203 \)[/tex]
- Standard deviation for Milwaukee: [tex]\( s_2 = 18.7426 \)[/tex]
3. Determine the Sample Sizes:
- Number of days sampled in New Orleans, [tex]\( n_1 = 20 \)[/tex]
- Number of days sampled in Milwaukee, [tex]\( n_2 = 25 \)[/tex]
4. Calculate the Test Statistic:
The test statistic for Welch's t-test is calculated using the formula:
[tex]\[
t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}
\][/tex]
Substituting the given values:
[tex]\[
t = \frac{63.5100 - 76.5480}{\sqrt{\frac{13.9203^2}{20} + \frac{18.7426^2}{25}}} = -2.6759
\][/tex]
5. Calculate the Degrees of Freedom:
The degrees of freedom for Welch's t-test is given by:
[tex]\[
df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1-1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2-1}}
\][/tex]
The calculated degrees of freedom is approximately:
[tex]\[
df \approx 42.8024
\][/tex]
6. Find the p-value:
The p-value is calculated using the cumulative distribution function (CDF) for the t-distribution:
[tex]\[
p \text{-value} = 2 \times \left(1 - \text{CDF}(|t|, df)\right)
\][/tex]
With [tex]\(t = -2.6759\)[/tex] and [tex]\(df = 42.8024\)[/tex], the p-value is approximately 0.0105.
7. Make the Decision:
Compare the p-value to the significance level [tex]\(\alpha = 0.10\)[/tex]:
- Since [tex]\( \text{p-value} = 0.0105 < \alpha = 0.10 \)[/tex], we reject the null hypothesis.
8. Conclusion:
The conclusion is that there is enough evidence at the 0.10 significance level to reject the null hypothesis. Therefore, we conclude that the mean temperature in New Orleans is different from the mean temperature in Milwaukee.
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