Questions: Listed below are the numbers of words spoken in a day by each member of eight different randomly selected couples. Complete parts (a) and (b) below. Male: 15,862, 25,747, 1,423, 7,714, 19,553, 15,773, 13,582, 26,500 Female: 24,505, 13,706, 18,910, 17,286, 13,012, 17,569, 16,334, 18,949 H1: μd ≠ 0 word(s) (Type integers or decimals. Do not round.) Identify the test statistic. t= (Round to two decimal places as needed.) Identify the P-value. P-value = (Round to three decimal places as needed.) What is the conclusion based on the hypothesis test? Since the P-value is greater than the significance level, fail to reject the null hypothesis. There is not sufficient evidence to support the claim that males speak fewer words in a day than females. b. Construct the confidence interval that could be used for the hypothesis test described in part (a). What feature of the confidence interval leads to the same conclusion reached in part (a)? The confidence interval is word(s) <μd< word(s). (Round to the nearest integer as needed.)

Listed below are the numbers of words spoken in a day by each member of eight different randomly selected couples. Complete parts (a) and (b) below. Male: 15,862, 25,747, 1,423, 7,714, 19,553, 15,773, 13,582, 26,500 Female: 24,505, 13,706, 18,910, 17,286, 13,012, 17,569, 16,334, 18,949 H1: μd ≠ 0 word(s) (Type integers or decimals. Do not round.) Identify the test statistic. t= (Round to two decimal places as needed.) Identify the P-value. P-value = (Round to three decimal places as needed.) What is the conclusion based on the hypothesis test? Since the P-value is greater than the significance level, fail to reject the null hypothesis. There is not sufficient evidence to support the claim that males speak fewer words in a day than females. b. Construct the confidence interval that could be used for the hypothesis test described in part (a). What feature of the confidence interval leads to the same conclusion reached in part (a)? The confidence interval is word(s) <μd< word(s). (Round to the nearest integer as needed.)
Transcript text: Listed below are the numbers of words spoken in a day by each member of eight different randomly selected couples. Complete parts (a) and (b) below. Male: 15,862, 25,747, 1,423, 7,714, 19,553, 15,773, 13,582, 26,500 Female: 24,505, 13,706, 18,910, 17,286, 13,012, 17,569, 16,334, 18,949 $\mathrm{H}_{1}: \mu_{\mathrm{d}}$ $\square$ 0 word(s) (Type integers or decimals. Do not round.) Identify the test statistic. $\mathrm{t}=$ $\square$ (Round to two decimal places as needed.) Identify the P-value. P-value $=$ $\square$ (Round to three decimal places as needed.) What is the conclusion based on the hypothesis test? Since the $P$-value is $\square$ greater than the significance level, fail to reject the null hypothesis. There is not sufficient evidence to support the claim that males speak fewer words in a day than females. b. Construct the confidence interval that could be used for the hypothesis test described in part (a). What feature of the confidence interval leads to the same conclusion reached in part (a)? The confidence interval is $\square$ word(s) $<\mu_{d}<$ $\square$ word(s). (Round to the nearest integer as needed.)
failed

Solution

failed
failed

Solution Steps

Step 1: Calculate the Standard Error

The standard error (SE) is calculated using the formula:

\[ SE = \frac{s_d}{\sqrt{n}} \]

where \(s_d = 10019.9492\) is the standard deviation of the differences and \(n = 8\) is the number of pairs. Thus,

\[ SE = \frac{10019.9492}{\sqrt{8}} \approx 3542.587 \]

Step 2: Calculate the Test Statistic

The test statistic \(t\) is calculated using the formula:

\[ t = \frac{\bar{d}}{SE} \]

where \(\bar{d} = -1764.625\) is the mean of the differences. Therefore,

\[ t = \frac{-1764.625}{3542.587} \approx -0.4981 \]

Step 3: Determine the Critical Value

For a two-tailed test at \(\alpha = 0.05\), the critical value \(t_{\alpha/2, df}\) is found using the t-distribution with \(df = n - 1 = 7\):

\[ t_{(0.025, 7)} \approx 2.3646 \]

Step 4: Calculate the P-value

The P-value is calculated as follows:

\[ P = 2 \times (1 - T(|t|)) = 2 \times (1 - T(0.4981)) \approx 0.6337 \]

Step 5: Conclusion of the Hypothesis Test

Since the P-value \(0.6337\) is greater than the significance level \(\alpha = 0.05\), we fail to reject the null hypothesis. This indicates that there is not sufficient evidence to support the claim that males speak fewer words in a day than females.

Step 6: Construct the Confidence Interval

The confidence interval for the mean difference \(\mu_d\) is calculated as:

\[ \text{Confidence Interval} = \bar{d} \pm t_{\alpha/2, df} \cdot SE \]

Calculating the bounds:

\[ \text{Lower Bound} = -1764.625 - 2.3646 \cdot 3542.587 \approx -10141 \] \[ \text{Upper Bound} = -1764.625 + 2.3646 \cdot 3542.587 \approx 6612 \]

Thus, the confidence interval is:

\[ -10141 < \mu_d < 6612 \]

Final Answer

The results of the hypothesis test and confidence interval are summarized as follows:

  • Test statistic \(t \approx -0.4981\)
  • P-value \( \approx 0.6337\)
  • Confidence interval: \(-10141 < \mu_d < 6612\)

\(\boxed{\text{Fail to reject the null hypothesis; confidence interval: } -10141 < \mu_d < 6612}\)

Was this solution helpful?
failed
Unhelpful
failed
Helpful

Questions asked by other users

failedAnswered
Find the inverse of (x=e^y).
failedAnswered
Intercalated disks are specialized connections only found in the following muscle(s): cardiac muscle skeletal muscle both none of the above
failedAnswered
Following is the probability distribution for age of a student at a certain public high school. x 13 14 15 16 17 18 P(x) 0.08 0.23 0.25 0.28 0.14 0.02 Part 1 of 2 (a) Find the variance of the ages. Round the answer to at least four decimal places. The variance of the ages is .
failedAnswered
(5^(-2))^(-5)
failedAnswered
Just output the content of the question, DO NOT output additional information or explanations.