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mathstatistics and probabilitystatistics and probability questions and answersconsider the following hypothesis statement using α=0.10 and data from two independent samples. assume the populaton variances are equai and the populafions are normaly distributed. complete parts a and b. h0:μ1−h2=0h1:μ1−μ2=0xˉ1=14.3s1=2.7n1=22.xˉ2=12.0s2=3.4n2=18 a. calculate the appeopriate sest statistic and interpret the result. the test statistic is
Question: Consider The Following Hypothesis Statement Using Α=0.10 And Data From Two Independent Samples. Assume The Populaton Variances Are Equai And The Populafions Are Normaly Distributed. Complete Parts A And B. H0:Μ1−H2=0H1:Μ1−Μ2=0xˉ1=14.3s1=2.7n1=22.Xˉ2=12.0s2=3.4n2=18 A. Calculate The Appeopriate Sest Statistic And Interpret The Result. The Test Statistic Is
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Consider the following hypothesis statement using α=0.10 and data from two independent samples. Assume the populaton variances are equai and the populafions are normaly distributed. Complete parts a and b.
H
0
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:μ
1
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−H
2
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=0
H
1
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:μ
1
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−μ
2
​


=0
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x
ˉ

1
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=14.3
s
1
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=2.7
n
1
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=22.
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x
ˉ

2
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=12.0
s
2
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=3.4
n
2
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=18
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a. Calculate the appeopriate sest statistic and interpret the result. The test statistic is (Round to two decimal places as needed.) The criscal value(s) is(are) (Round to two decimal places as needed. Uso a comma to soparate answers as needed.) Because the test statissic the null hypothesis b. Identity the p-value from part a and inserpret the result. Thep-value is (Round to three decimal places as needed.) Interptet the result. Choose the correct answer below. A. Since the p-value is less than the significance ievel, reject the null hypothesis. B. Since the p-value is not less than the signficance lovel, reject the null hypothesis. c. Since the p-value is not less than the significance level, do not ieject the nuit hypothesis D. Since the ovalue is less than the significance level, do not reject the nuil hypothesis.

Answer 1

To calculate the appropriate test statistic, we can use the formula for the two-sample t-test:

\[ t = \frac{{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}}{{\sqrt{\frac{{s_1^2}}{{n_1}} + \frac{{s_2^2}}{{n_2}}}}} \]

Given:

\(\bar{x}_1 = 14.3\), \(s_1 = 2.7\), \(n_1 = 22\)

\(\bar{x}_2 = 12.0\), \(s_2 = 3.4\), \(n_2 = 18\)

Assuming equal variances and normally distributed populations, we can proceed with the calculations.

First, let's calculate the standard error:

\[ \text{Standard Error} = \sqrt{\frac{{s_1^2}}{{n_1}} + \frac{{s_2^2}}{{n_2}}} \]

\[ = \sqrt{\frac{{2.7^2}}{{22}} + \frac{{3.4^2}}{{18}}} \]

\[ \approx 0.933 \]

Now, substitute the values into the formula to calculate the test statistic:

\[ t = \frac{{(14.3 - 12.0) - 0}}{{0.933}} \]

\[ \approx 2.464 \]

The test statistic is approximately 2.464.

Next, to interpret the result, we need to compare the test statistic with the critical value(s). The critical value(s) depend on the significance level (α) and the degrees of freedom (df). Since we are using α = 0.10, we need to find the critical value(s) that correspond to a one-tailed test with a 10% level of significance.

Using a t-distribution table or a statistical software, we find that for a one-tailed test with α = 0.10 and df = 22 + 18 - 2 = 38, the critical value is approximately 1.311.

Comparing the test statistic (2.464) with the critical value (1.311), we can see that the test statistic is greater than the critical value. This means that the test statistic falls into the rejection region.

b. To determine the p-value, we need to find the area under the t-distribution curve that is more extreme than the observed test statistic. In this case, since it is a one-tailed test with Ha: μ1 - μ2 ≠ 0, we need to find the p-value for the right-tail.

Using a t-distribution table or a statistical software, we find that the p-value is less than 0.05 (rounding to three decimal places).

Interpreting the result, we can conclude that the p-value is less than the significance level (α = 0.10). Therefore, we reject the null hypothesis (H0) and have sufficient evidence to support the alternative hypothesis (Ha). This indicates that there is evidence to suggest that the mean time of Team 1 is significantly different from the mean time of Team 2.