Question

The advertised diameter of steel rods manufactured by a certain company is 50 mm. During a routine check, five rods are selected randomly from a large lot and their diameters are measured yielding the following values: 50.16, 50.27, 50.13, 50.22, 50.32. The diameter can be assumed to be at least approximately normally distributed. At a 5% significance level, does the evidence suggest that the diameter of rods manufactured by the company is not 50 mm? Report also a p-value.

Answer 1

First enter Data into EXCEL

We have to find the sample mean.

Excel command is =AVERAGE(Select data)

sample mean =

Now we have to find sample standard deviation.

Excel command is =STDEV(Select data)

standard deviation = s = 0.078

n = 5

S =0.078

claim : The diameter of rods manufactured by the company is not 50 mm

Null and alternative hypothesis is

H0 : u = 50

H1 : u ≠ 50

Level of significance = 0.05

Here population standard deviation is not known so we use t-test statistic.

Test statistic is

Degrees of freedom = n - 1 = 5 - 1 = 4

P-value = 0.0032 ( using t table)

P-value , Reject H0

conclusion : At a 5% significance level, does the evidence suggest that the diameter of rods manufactured by the company is not 50 mm