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ivanzaharov [21]
3 years ago
6

A simple random sample with n = 58 provided a sample mean of 24.5 and a sample standard deviation of 4.4. (Round your answers to

one decimal place.) (a) Develop a 90% confidence interval for the population mean. to
Mathematics
1 answer:
nasty-shy [4]3 years ago
6 0

Answer: (23.53, 25.47)

Step-by-step explanation:

The confidence interval for population mean(when population standard deviation is unknown) is given by :-

\overline{x} \pm\ t^*\dfrac{s}{\sqrt{n}}

, where \overline{x} = sample mean

n = sample size.

s = sample standard deviation.

t* = Critical value.

Given : \overline{x}=24.5

s= 4.4

Confidence level = 90% =0.090

Significance level = \alpha=1-0.90=0.10

Sample size : n= 58

Degree of freedom : df= n-1= 57

Using t-distribution table , the critical value :

t_{\alpha/2,\ df}= t_{0.05,\ 57}=1.6720

Then, the confidence interval will be :-

24.5 \pm\ (1.6720)\dfrac{4.4}{\sqrt{58}}

24.5 \pm\ (1.6720)\dfrac{4.4}{7.61577310586}

24.5 \pm\ 0.965995165263\approx24.5\pm0.97\\\\=(24.5-0.97,\ 24.5+0.97)\\\\=(23.53,\ 25.47)

Hence, a 90% confidence interval for the population mean = (23.53, 25.47)

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