A random sample of 160 pumpkins is obtained and the mean circumference is found

Question

A random sample of 160 pumpkins is obtained and the mean circumference is found to be 46 cm. Assuming that the population standard deviation is known to be 1.5 cm, use a 0.01 significance level to test the claim that the mean circumference of all pumpkins is equal to 40.00 cm. A random sample of 160 pumpkins is obtained and the mean circumference is found to be 46 cm. Assuming that the population standard deviation is known to be 1.5 cm, use a 0.01 significance level to test the claim that the mean circumference of all pumpkins is equal to 40.00 cm.

Explanation / Answer

Solution:

Here, we have to use one sample z test for the population mean. The null and alternative hypotheses for this test are given s below:

Null hypothesis: H0: The mean circumference of all pumpkins is equal to 40.00 cm.

Alternative hypothesis: Ha: The mean circumference of all pumpkins is not equal to 40.00 cm.

H0: µ = 40 versus Ha: µ 40

This is a two tailed test.

We are given

Sample size = n = 160

Sample mean = Xbar = 46

Population standard deviation = = 1.5

Level of significance = = 0.01

The test statistic formula is given as below:

Z = (Xbar - µ) / [/sqrt(n)]

Z = (46 – 40) / [1.5/sqrt(160)]

Z = 6/0.118585

Z = 50.59644

Critical values = -2.5758 and 2.5758 (By using z-table or excel)

P-value = 0.00 (By using z-table or excel)

= 0.01

P-value < = 0.01

So, we reject the null hypothesis that the mean circumference of all pumpkins is equal to 40.00 cm.

There is insufficient evidence to conclude that the mean circumference of all pumpkins is equal to 40.00 cm.

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