How can the following be evaluated using excel? A tire company makes 5000 tires

Question

How can the following be evaluated using excel?

A tire company makes 5000 tires an hour on average with a known standard deviation of 50. A sample is taken last hour and 5100 tires were made. Spell out the null and alternate hypotheses that the mean is different than 5000. Evaluate the hypothesis (Accept or Reject) at the following significance levels:

0.01

0.05

Samples of a car company's output is taken and 490 cars are made in a day with a known standard deviation of 16 cars. Spell out the null and alternate hypotheses that the average production of the car company is less than 475 cars. Evaluate the hypothesis (Accept or Reject) at the following significance levels?:

0.001

0.1

Explanation / Answer

A tire company makes 5000 tires an hour on average with a known standard deviation of 50. A sample is taken last hour and 5100 tires were made. Spell out the null and alternate hypotheses that the mean is different than 5000. Evaluate the hypothesis (Accept or Reject) at the following significance levels:
0.01
0.05

Formulating the null and alternative hypotheses,              
              
Ho:   u   =   5000  
Ha:    u   =/   5000  
              
As we can see, this is a    two   tailed test.      
              
Getting the test statistic, as              
              
X = sample mean =    5100          
uo = hypothesized mean =    5000          
n = sample size =    1 [just the sample taken last hour]          
s = standard deviation =    50          
              
Thus, z = (X - uo) * sqrt(n) / s =    2          
              
Also, the p value is              
              
p =    0.045500264          

[You can get this P value by typing

=2*(1-NORMSDIST(2))

on any empty cell. The 2 at the beginning is becasue it is two tailed. 1 - NORMSDIST(2) give the right tailed area for z = 2, as NORMSDIST(2) itself is the left tailed area of z = 2.]
              
As P > 0.01, we FAIL TO   REJECT THE NULL HYPOTHESIS at 0.01 level.

As P < 0.05, we REJECT THE NULL HYPOTHESIS at 0.05 level.

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