Question

An engineer measures the weights (in kilograms) of steel pieces. They would like to
test H0 : = 5 against H1 : > 5. The weight of a steel piece is normally distributed.
They select a random sample size of n = 25 steel pieces, and compute x = 6:7 and
s = 2:37. We cannot conclude that the mean weight is larger than 5 kg at a level of
significance of 1%.
True False

An engineer measures the weights (in kilograms) of steel pieces. They would like to test H. := 5 against H1 :pl > 5. The weight of a steel piece is normally distributed. They select a random sample size of n = 25 steel pieces, and compute T = 6.7 and 2.37. We cannot conclude that the mean weight is larger than 5 kg at a level of significance of 1%. S = True False

Answer 1

Answer:

False

Solution:

Here, we have to use one sample t test for the population mean.

The null and alternative hypotheses are given as below:

H₀: µ = 5 versus H_a: µ > 5

This is an upper tailed test.

The test statistic formula is given as below:

t = (Xbar - µ)/[S/sqrt(n)]

From given data, we have

µ = 5

Xbar = 6.7

S = 2.37

n = 25

t = (Xbar - µ)/[S/sqrt(n)]

t = (6.7 - 5)/[2.37/sqrt(25)]

t = 3.5865

df = n – 1 = 25 - 1 = 24

α = 0.01

Critical value = 2.4922

(by using t-table or excel)

P-value = 0.0007

(by using t-table)

P-value < α = 0.01

So, we reject the null hypothesis

There is sufficient evidence to conclude that the mean weight is larger than 5 kg at a level of significance of 1%.

So, given statement is False.