Question

Mean number of desks produced per week is 42 and population standard deviation is 4.67. the company has introduced new production methods. A random sample of 12 weeks production indicates 44 desks were produced each week. has the introduction of new production methods increased average number of desks produced each week at .05 significance level. Estimate the 95% confidence interval

Answer 1

Solution:

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

H₀: µ = 42

H_a: µ > 42

This is an upper tailed test.

We are given

Level of significance = α = 0.05

Test statistic formula is given as below:

Z = (Xbar - µ)/[σ/sqrt(n)]

We are given

Xbar = 44

µ = 42

σ = 4.67

n = 12

Z = (44 – 42)/[4.67/sqrt(12)]

Z = 1.4836

P-value = 0.0690

(by using z-table)

P-value > α = 0.05

So, we do not reject the null hypothesis

There is insufficient evidence to conclude that introduction of new production methods increased average number of desks produced each week.

Now, we have to find the 95% confidence interval.

Confidence interval for Population mean is given as below:

Confidence interval = Xbar ± Z*σ/sqrt(n)

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

Xbar = 44

σ = 4.67

n = 12

Confidence interval = Xbar ± Z*σ/sqrt(n)

Confidence interval = 44 ± 1.96*4.67/sqrt(12)

Confidence interval = 44 ± 1.96*1.3481

Confidence interval = 44 ± 2.6423

Lower limit = 44 - 2.6423 = 41.3577

Upper limit = 44 + 2.6423 = 46.6423

Lower limit = 41.3577

Upper limit = 46.6423