Question

A car company executive claims that a mean of 48.3 cars per dealership are being sold each month. An important stockholder thinks this claim is high and runs a test by sampling 30 dealerships. What conclusion can be reached of the sample mean is 45.4 cars with a standard deviation of 15.4? There is sufficient evidence to prove the executive's claim is true. There is sufficient evidence to prove the executive's claim is false. The stockholder has sufficient evidence to reject the executive's claim. O The stockholder does not have sufficient evidence to reject the executive's claim. There is not sufficient data to reach any conclusion.

Answer 1

Solution:

Claim by a executive ; $\mu$ = 48.3

Stockholder think that  $\mu$ <  48.3

So , the null and alternative hypothesis can be written as

H₀ :  $\mu$ = 48.3 vs H₁ :  $\mu$ <  48.3

Given ,

n = 30

$\bar x$ = 45.4

s = 15.4

Since population SD is unknown,we use t test.

The test statistics t is given by ..

t =  $url=https://latex.codecogs.com/gif.latex?%5Cfrac%7B%5Cbar%20x-%5Cmu%7D%7Bs/%5Csqrt%7Bn%7D%7D&t=1583903773&ymreqid=f4764063-0d40-e2b0-1c91-4d0044010500&sig=YUWl_kKRL7NxSivf248T.Q--~C$

= (45.4 - 48.3)/(15.4/$url=https://latex.codecogs.com/gif.latex?%5Csqrt%7B%7D&t=1583903773&ymreqid=f4764063-0d40-e2b0-1c91-4d0044010500&sig=JpAAxOZN566tRfgdlnWZJA--~C$30)

= -1.031

t = -1.031

Left tailed test

d.f. = n - 1 = 29

So , using calculator

p value = 0.155

We generally take the level of significance at 0.01 or 0.05 or 0.10

p value > significance level.

So , we FAIL TO REJECT H₀

The stockholder does not have the sufficient evidence to reject the executive's claim.