Question

This Question: 1 pt A random sample of 40 adults with no children under the age of 18 years results in a mandaly sure time of 5.56 hours, with a standard deviation of 2.38 hours. A random sample of nuts with children under the age of 18 rute name dalybere ne of 4.32 hours, with standard deviation of 1.53 houn. Construct and interpreta % confidence interval for the mean difference in late time between adults with no children and adults with children (A) Let per represent the mean leisure hours of us with no children under the age of 18 and represent the mean leisure hours of us with children under the age of 16. The confidence tarvetter (P4-22) the metromhours to hous (Round to two decimal places as needed) What is the interpretation of this confidence interval OA. There is a probably that the difference of the means is in the Werval Conclude that there is insufficient evidence of a significant difference in the number of our hours OB. There is one that the difference of the means in in the interval Conclude that there is nofciant evidence de cantorence in the number of wire hours OC. There is a probably that the difference of the means is in the interval conclude that there is a significant difference in the number of our hour OD. There is one that the ofference of the means is in the terve. Conclude that there is a great diference in the number of Warehoun Click to set your news MacBook Air Po

Answer 1

Adults with no children,

number of samples, n= 40

$Mean, \mu_1 = 5.55 \; hours$

$standard\; deviation,\sigma =2.38 \;hours$

Adults with children

number of samples, n= 40

$Mean, \mu_2 = 4.32 \; hours$

$standard\; deviation,\sigma =1.53 \;hours$

The 95% confidence interval for $(\mu_1 - \mu_2)$ is given by,

$Confidence\;interval = (\mu_1-\mu_2)\pm z*\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$

$Confidence\;interval = (5.55-4.32)\pm 1.96*\sqrt{\frac{2.38^2}{40}+\frac{1.53^2}{40}}$

Therefore,

The 95% confidence interval for $(\mu_1 - \mu_2)$ is the range from 0.35 hours to 2.21 hours

The interpretation is,

D. There is 95% confidence that the diffrence of the means is in the interval. Conclude that there is a significant difference in the number of leisure hours.

Reason : Since the confidence interval does not include 0 , we can conclude that there is significant difference in the means.