Question

A manufacturer of potato chips would like to know whether its bag filling machine works correctly at the 409.0 gram setting. Based on a 36 bag sample where the mean is 414.0 grams, is there sufficient evidence at the 0.05 level that the bags are overfilled? Assume the standard deviation is known to be 26.0.

Step 1: Enter the hypotheses:

Ho:___________________

Ha:___________________

Step 2: Find the value of the test statistic. Round your value to three decimal places

Step 3: Specify if the test is one tailed or two tailed

Step 4: Determine the decision rule for rejecting the null hypothesis. Round your value to three decimal places.

Step 5. Enter the value of the level of significance.

Step 6: Make the decision to reject or fail to reject the Null hypothesis.

(A) Reject the null hypothesis

(B) Fail to reject the null hypothesis

Answer 1

Solution :

= 409.0

= 414.0

= 26.0

n = 36

This is the two tailed test .

The null and alternative hypothesis is ,

H₀ :   = 409

H_a :    409

Test statistic = z

= ( - ) / / n

= (414 - 409) /26.0 / 36

= 1.154

p(Z > 1.154 ) = 1-P (Z <1.154 ) =0.2486

P-value = 0.2486

= 0.05  

p= 0.2486 ≥ 0.05, it is concluded that the null hypothesis is not rejected.

Reject the null hypothesis .

The significance level is α=0.05, and the critical value for a two-tailed test is z_c​=1.96.

There is not enough evidence to claim that the population mean μ is different than 409, at the 0.05 significance level.