Question

The melting point (in degrees Fahrenheit) for a randomly selected tub of margarine of a certain brand is known to have a normal distribution with a SD, σ = 1.21. A sample of n = 5 tubs of margarine is drawn. The sample mean of melting temperature is computed to be ¯x = 95.3.

(a) Construct a 95% confidence interval for the mean melting temperature.

(b) Construct a 99% confidence interval for the mean melting temperature. Compare the width of the two intervals in (a) and (b)?

(c) Did you use the central limit theorem to construct the intervals in (a) and (b)? Explain.

Answer 1

Solution-A:

z crit fro 95%=1.96

95% confidence interval for mean is

xbar-z*sigma/sqrt(n),xbar+z*sigma/sqrt(n)

95.3-1.96*1.21/sqrt(5),95.3+1.96*1.21/sqrt(5)

94.23939, 96.36061

95% lower limir mean=94.23939

95% upper limir mean=96.36061

Solution-b:

z crit fro 99%=2.576

99% confidence interval for mean is

xbar-z*sigma/sqrt(n),xbar+z*sigma/sqrt(n)

95.3-2.576*1.21/sqrt(5),95.3+2.576*1.21/sqrt(5)

93.90605, 96.69395

99% lower limir mean=93.90605

99% upper limir mean=96.69395

99% interval is wider than 95%

(c) Did you use the central limit theorem to construct the intervals in (a) and (b)? Explain.

yes we used central limit theorem ,as the original sample follows normal distribution,sampling distribution follows normal distribution.

sample mean=xbar=mu=95.3

sample standard deviation=sigma/sqrt(n)=1.21/sqrt(5)