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Wewaii [24]
3 years ago
14

Using the simple random sample of weights of women from a data​ set, we obtain these sample​ statistics: nequals45 and x overbar

equals148.79 lb. Research from other sources suggests that the population of weights of women has a standard deviation given by sigmaequals31.37 lb. a. Find the best point estimate of the mean weight of all women. b. Find a 90​% confidence interval estimate of the mean weight of all women.
Mathematics
1 answer:
Tomtit [17]3 years ago
3 0

Answer: (141.1, 156.48)

Step-by-step explanation:

Given sample statistics : n=45

\overline{x}=148.79\text{ lb}

\sigma=31.37\text{ lb}

a) We know that the best point estimate of the population mean is the sample mean.

Therefore, the best point estimate of the mean weight of all women = \mu=148.79\text{ lb}

b) The confidence interval for the population mean is given by :-

\mu\ \pm E, where E is the margin of error.

Formula for Margin of error :-

z_{\alpha/2}\times\dfrac{\sigma}{\sqrt{n}}

Given : Significance level : \alpha=1-0.90=0.1

Critical value : z_{\alpha/2}=z_{0.05}=\pm1.645

Margin of error : E=1.645\times\dfrac{31.37}{\sqrt{45}}\approx 7.69

Now, the 90% confidence interval for the population mean will be :-

148.79\ \pm\ 7.69 =(148.79-7.69\ ,\ 148.79+7.69)=(141.1,\ 156.48)

Hence, the 90​% confidence interval estimate of the mean weight of all women= (141.1, 156.48)

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