A random sample of 100 pumpkins is obtained and the mean circumference is found to be 40.5 cm. Assuming the population standard deviation is known to be 1.6cm, use a 0.05 significance level to test the claim that the mean circumference is equal to 39.9 cm.

We are given that a random sample of 100 pumpkins is obtained and the mean circumference is found to be 40.5 cm and assuming the population standard deviation is known to be 1.6 cm.

Let Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 39.9 cm {where [tex]\mu[/tex] = population mean circumference}

Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu\neq[/tex] 39.9 cm

The test statistics we will use here is;

T.S. = [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)

where, Xbar = sample mean circumference = 40.5 cm

[tex]\sigma[/tex] = population standard deviation = 1.6 cm

n = sample of pumpkins = 100

So, test statistics = [tex]\frac{40.5-39.9}{\frac{1.6}{\sqrt{100} } }[/tex]

= 3.75

Now, at 0.05 significance level z table gives critical value of 1.96. Since our test statistics is more than the critical value of z which means test statistics will lie in the rejection region, so we have sufficient evidence to reject null hypothesis.

Therefore, we conclude that the population mean circumference is different from 39.9 cm .