Adults in Turkey have mean caloric intake of 3500 calories per day (Wikipedia) with a standard deviation of 500 calories. A sample of 25 Ankara University students has an average intake of 3400 calories per day. Is this enough evidence to conclude that Ankara U. students average is not 3500 calories per day? Assume the distribution is normal. A. State the null and alternative hypotheses.
B. What is the value of the test statistic?
C. Specify the rejection region?
D. State the conclusion.

A) Null hypothesis; H0: μ = 3500

Alternative hypothesis;Ha μ ≠ 3500

B) t = -1

C) the rejection region will be;

p-value < 0.05

D) Conclusion is that there is no sufficient evidence to support the claim that Ankara U. students average is not 3500 calories per day.

Step-by-step explanation:

We are given;

Population mean; μ = 3500

Population standard deviation; σ = 500

Sample mean; x¯ = 3400

Sample size; n = 25

We want to find enough evidence to conclude that Ankara U. students average is not 3500 calories per day.

Thus, the hypothesis is defined as;

A) Null hypothesis; H0: μ = 3500

Alternative hypothesis;Ha μ ≠ 3500

B) sample size is less than 30, thus we will use a t-test.

Formula for the test statistic is;

t = (x¯ - μ)/(σ/√n)

t = (3400 - 3500)/(500/√25)

t = -100/(500/5)

t = -1

C) since the sample is small, let's assume a significance level of 0.05.

Thus,the rejection region will be;

p-value < 0.05

D) from online p-value from t-score calculator attached using;

t = - 1; DF = 25 - 1 = 24; significance level = 0.05 and two tailed hypothesis, we have;

p-value ≈ 0.3273

This is greater than the significance value of 0.05.

Thus, we will fail to reject the null hypothesis and conclude that there is no sufficient evidence to support the claim that Ankara U. students average is not 3500 calories per day.

if a = 1,   and b = 2, then the answer is 47.99999999. if you round that to the nearest whole number, the answer is 48.

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