With a mean of 100 and standard deviation of 15, what percentage of IQ scores are less than 110

A national grocery store chain wants to test the difference in the average weight of turkeys sold in Detroit and the average wei

Arte-miy333 [17]

Answer:

Calculated value t = 1.3622 < 2.081 at 0.05 level of significance with 42 degrees of freedom

The null hypothesis is accepted .

Assume the population variances are approximately the same

Step-by-step explanation:

Explanation:-

Given data a random sample of 20 turkeys sold at the chain's stores in Detroit yielded a sample mean of 17.53 pounds, with a sample standard deviation of 3.2 pounds

The first sample size 'n₁'= 20

mean of the first sample 'x₁⁻'= 17.53 pounds

standard deviation of first sample S₁ = 3.2 pounds

Given data a random sample of 24 turkeys sold at the chain's stores in Charlotte yielded a sample mean of 14.89 pounds, with a sample standard deviation of 2.7 pounds

The second sample size n₂ = 24

mean of the second sample "x₂⁻"= 14.89 pounds

standard deviation of second sample S₂ = 2.7 pounds

Null hypothesis:-H₀: The Population Variance are approximately same

Alternatively hypothesis: H₁:The Population Variance are approximately same

Level of significance ∝ =0.05

Degrees of freedom ν = n₁ +n₂ -2 =20+24-2 = 42

Test statistic :-

$t = \frac{x^{-} _{1} - x_{2} }{\sqrt{S^2(\frac{1}{n_{1} } }+\frac{1}{n_{2} } }$