Question

In a survey sample of 83 respondents, the mean number of hours worked per week is 39.04, with a standard deviation of 11.51. Calculate a 95 percent confidence interval for these data, and interpret your result.

Answer 1

Given that,

Sample size= 83

Mean number= 39.04

Standard deviation= 11.51

We know the critical t-value for  95% confidence interval which is equal to 1.989.

We also know the formula for confidence interval,

CI=( mean number - critical t-value*standard deviation/(sample size)^(1/2), mean number + critical t-value*standard deviation/(sample size)^(1/2))

So, we have

CI= (39.04 - 1.989*11.51/83^(1/2), 39.04 + 1.989*11.51/83^(1/2)

CI= (39.04 - 2.513,39.04 + 2.513)

CI= (36.527,41.553)

Therefore, 95% confidence interval for these data is (36.527,41.553), and this result interpret that the true value for this survey sample lie in the interval (36.527,41.553).

Answer 2

Answer: We are 95% confidence that the respondents  are worked between 36.57 hours and 41.51 hours.

Explanation:

Since we have given that

Number of respondents in a survey sample = 83

Mean number of hours worked per week = 39.04

Standard deviation = 11.51

Now, we'll calculate standard error of mean is given by

$\sigma_\bar{x}=\frac{\sigma}{\sqrt{n}}=\frac{11.51}{\sqrt{83}}=1.26$

95% of confidence interval is given by

$\bar{x}\pm 1.96\sigma_{\bar{x}}\\\\39.04\pm 1.96\times 1.26\\\\39.04\pm 2.47\\\\\{41.51,36.57\}$

Hence, we are 95% confidence that the respondents  are worked between 36.57 hours and 41.51 hours.