5. A nutrition researcher undertakes a study concerning steroid content in paste

Question

5. A nutrition researcher undertakes a study concerning steroid content in pasteurized milk. Based on his research, the researcher concludes that the population mean steroid content per quart of milk is 893 g. Another researcher, upon hearing this result, becomes skeptical of this claim and decides to test it himself. He draws a sample with six observations (six quarts of milk) each with steroid content as follows: Quart # Mean steroid content (in 4 890 902 947 877 911 881 Define the random variable X to be the steroid content (in g) of a randomly selected quart of The null hypothesis is that the expected steroid content (in g) of a randomly selected quart of milk is 893 ug. The research (alternate) hypothesis is that that the expected steroid content (in g) of a randomly selected quart of milk exceeds 893 g. Set up the appropriate null and alternate hypotheses a. Calculate the P-value for the test. Based on the P-value only, do you accept or reject the null hypothesis? Why? b. Construct the appropriate confidence interval for the hypothesis test of part a) with P(U)-.01. Using the confidence interval only, do you accept or reject the null hypothesis? Why? Are any assumptions on the random variable X required to guarantee that your answers to parts a) and b) are valid? c.

Explanation / Answer

a)

Below are the null and alternate hypothesis

H0: mu <= 893
H1: mu > 893

This is right tailed test.
n = 6
xbar = 901.3333
std. dev. s = 25.7423

Test statistics, t = (xbar - mu)/(s/sqrt(n))
t = (901.333 - 893)/(25.7423/sqrt(6))
t = 0.7929

df = 6 - 1 = 5

p-value = 0.2319
p-value is calculated using standard t-table or excel formula : =1 - T.DIST(0.7929,5,TRUE))

As p-value is greater than the significance level of 0.05, we fail to reject the null hypothesis.
There are not significant evidence to conclude that average steroid content is greater than 893.

b)

As confidence interval includes 893, we fail to reject the null hypothesis.

c)

Assumption that the data follows normal rule.

CI for 99% n 6 mean 901.3333 t-value of 99% CI 4.0321 std. dev. 25.7423 SE = std.dev./sqrt(n) 10.50925 ME = t*SE 42.37480 Lower Limit = Mean - ME 858.95850 Upper Limit = Mean + ME 943.70810 99% CI (858.9585 , 943.7081 )

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