Question

For each of the following, please show the following to receive full credit. (Answers without these steps will not be given credit)

(a) Draw a curve and label each part to reflect a left, right, or two-tailed test

(b) Claim: Show in symbols H0, Ha

(c) Critical value- find the z0 or t0 values

(d) What are the standardized test statistic z or t, and the P values from the calculator?

(e) Comparing the critical values, α value, against the calculator results, state whether you reject, or fail to reject the null hypothesis

(f) Interpret the decision in the context of the original claim.

--A consumer group claims that the mean annual consumption of high fructose corn syrup by a person in the U.S. is 48.8 pounds. A random sample of 120 people in the U.S. has a mean annual high fructose corn syrup consumption of 49.5 pounds. Assume the population standard deviation is 3.6 pounds. At α = 0.05, can you reject the claim?--

Answer 1

Let $\mu$ denotes the mean annual consumption of high fructose corn syrup by a person in the U.S.

This test is a two tailed test as alternative hypothesis contains $\neq$ sign,

Hypotheses : To test null hypothesis H. :u=48.8 against alternative hypothesis H1 :p # 48.8 Let i denotes the sample mean, o denotes the population standard deviation and n denotes the sample size. Here, T = 49.5000, o2 = 12.9600, 0 = 3.60, n = 120 The test statistic can be written as: Vnī – 48.8) which under H, follows a standard normal distribution. Decision rule / Rejection region : We reject Ho at 0.05 level of significance if P-value < 0.05 or if Zobs > 20.025 = 1.959964 0.4- 0.3 0.2 0.1 - 0.0 - -4 -2 Z Now, Value of test statistic: The value of the test statistic is Zobs V120(49.5000 – 48.8) 3.60 2.130032 2.13 The critical value =+ 1.959964 +1.960 P-value = P(Z > Zobs) = 2 * P(Z > 2.130032) = 2 * (1 – 0.983416) = 0.033169 0.0332 Conclusion : Since p-value < 0.05 and zobs > Zcritical = 1.96, so we reject H, at 0.05 level of significance Hence, we can conclude that the the population mean is not equal to 48.8.