Question

A new baker is trying to decide if he has an appropriate price set for his 3 tier wedding cakes which he sells for $88.97. He is particullarly interested in seeing if his wedding cakes sell for less than the average price. He searches online and finds that out of 30 of the competitors in his area they sell their 3 tier wedding cakes for $86.78. From a previous study he knows the standard deviation is $5.79. Help the new baker by testing this with a 0.05 level of significance.

The correct hypotheses would be:

• H0:μ≤$88.97H0:μ≤$88.97
  HA:μ>$88.97HA:μ>$88.97 (claim)
• H0:μ≥$88.97H0:μ≥$88.97
  HA:μ<$88.97HA:μ<$88.97 (claim)
• H0:μ=$88.97H0:μ=$88.97
  HA:μ≠$88.97HA:μ≠$88.97 (claim)

Since the level of significance is 0.05 the critical value is 1.645

The test statistic is: (round to 3 places)

The p-value is: (round to 3 places)

The decision can be made to:

• reject H0H0
• do not reject H0H0

The final conclusion is that:

• There is enough evidence to reject the claim that his wedding cakes sell for less than the average price.
• There is not enough evidence to reject the claim that his wedding cakes sell for less than the average price.
• There is enough evidence to support the claim that his wedding cakes sell for less than the average price.
• There is not enough evidence to support the claim that his wedding cakes sell for less than the average price.

Answer 1

x̅ = 86.78, σ = 5.79, n = 30  

α = 0.05

Null and Alternative hypothesis:

Ho : µ ≥ 88.97  

H1 : µ < 88.97 (claim)

Critical value :   

Left tailed critical value, z crit = NORM.S.INV(0.05) = -1.645

Test statistic:   

z = (x̅- µ)/(σ/√n) = (86.78 - 88.97)/(5.79/√30) = -2.072

p-value :   

p-value = NORM.S.DIST(-2.072, 1) = 0.0191

Decision:   

p-value < α, Reject the null hypothesis.   

Conclusion:

There is enough evidence to support the claim that his wedding cakes sell for less than the average price.