Question

A manufacturer claims that the yhickness of the spearmint gum it produces is 7.5 one-hundredths of an inch. A quality control specialist reqularly checks this claim. On one production run,he took a random sample n=10 pieces of gum and measured their thickness. He obtained as follows. Construct and test hypothesis at %5 significance level.

7.65	7.60	7.65	7.70	7.55
7.55	7.40	7.40	7.50	7.50

Answer 1

Null hypothesis  

$H_{0}:\mu =7.5$

vs
                  
Alternative hypothesis

$H_{1}:\mu \neq 7.5$                  
                  
We have for given sample,                  
Population mean for given sample=7.5  
Sample mean=7.55          
Sample standard deviation s=0.1027      
Sample size =n=10          
                  
Degree of freedom = n-1 =9      
t test statistic formula is

$t=\frac{\bar{x}-\mu }{\frac{s}{\sqrt{n}}}=\frac{7.55-7.5 }{\frac{0.1027}{\sqrt{10}}} =1.54$

P value is 0.1580...............................by using TDIST(1.54,9,2) in Excel.

P value is 0.1580 > 0.05

Therefore , we fail to reject H0.