Question

Q.1
a. Find a 90% confidence interval for the mean of normal distribution if σ =3 and if a sample of size 10 gave the values 9, 14, 10, 12, 7, 13, 11, 12, 15, 12. What would be the confidence interval if σ is unknown?
b. The breaking strength of 10 specimens of 0.104 inches diameter hard drawn copper wire are found to be 578, 572, 570, 568, 572, 570, 570, 572, 596, 584 pounds. Calculate the 95% confidence interval for the mean breaking strength of this kind of wire, assuming that the measurements are normally distributed. Also test the hypothesis that the average breaking strength of this type of wire is 590.

Answer 1

a) since the population standard deviation is known we use the z distribution.

Obtain the sample mean.

90% CI=$\bar x ±z_{\frac{\alpha} {2} }\frac{\sigma}{\sqrt n}$

=$11.5±1.645*\frac{3}{\sqrt{10}}$

=(9. 9394, 13.0606)

If the population standard deviation is unknown, we use the t distribution.

Calculate the sample standard deviation.

90% CI= $\bar x ±t_{({\frac{\alpha} {2}, n-1)}} \frac{s}{\sqrt n}$

=$11.5±1.833*\frac{2.3688}{\sqrt{10}}$

(10.1269, 12.8731)

b) obtain the sample mean and standard deviation.

95%CI=$\bar x ±t_{({\frac{\alpha} {2}, n-1)}} \frac{s}{\sqrt n}$

$=575.2±2.262*\frac{8.7025}{\sqrt{10}}$

=(568. 975, 581.425)

We test the hypotheses

H0: average breaking strength is 590

H1: average breaking strength is not equal to 590.

Since 590 is not within the 95%confidence interval range, we reject our null hypothesis in favor of the alternative. The average breaking strength is not equal to 590 at 5% level of significance.