Let [tex]\mu[/tex] denotes the average width of stripes .
As per given , we have
[tex]H_0:\mu=4\\H_a:\mu\neq4[/tex]
, since [tex]H_a[/tex] is two-tailed , so the test is a two-tailed test.
Also, population standard deviation is unknown , so we perform two-tailed t-test.
For Sample size : n= 45
Sample mean : [tex]\overline{x}=3.87[/tex]
Sample standard deviation : s= 0.5 inches
Test statistic : [tex]t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}[/tex]
i.e. [tex]t=\dfrac{3.87-4}{\dfrac{0.5}{\sqrt{45}}}\approx-1.74[/tex]
Significance level = [tex]\alpha=0.05[/tex]
By using t-value table,
Two-tailed critical t-value = [tex]t_{\alpha/2,df}=t_{0.025,\ 44}=\pm2.0154[/tex] [df = n-1]
Decision : Since the test statistic value (-1.74) lies with in the interval (-2.0154, 2.0154) , it means we are failed to reject the null hypothesis .
Conclusion: We have sufficient evidence to support the claim that the machine has slipped out of adjustment and the average width of stripes is no longer μ = 4 inches.
