Question

Question 1  

A single random sample of 16 different cereals is obtained, and the sugar content (in grams of sugar per gram of cereal) is measured for each cereal selected. Those amounts have a mean of 0.295 gram and a standard deviation of 0.168 gram. The amount of sugar in all cereals is assumed to be normally distributed.

Construct a 95% confidence interval estimate of the population mean.

Degrees of freedom =                            [ Select ]                       ["14", "15", "16", "17"]      

t-value =                            [ Select ]                       ["2.131", "2.120", "1.753", "1.746"]         (from the table)

Margin of Error =                            [ Select ]                       ["0.0357", "0.0584", "0.0895", "0.168"]      

95% Confidence Interval =                            [ Select ]                       ["0.2722 < μ < 0.3845", "0.2055 < μ < 0.3845", "0.2055 < μ < 0.3243", "0.2722 < μ < 0.3243"]      

--------------------------------------------

Question 2  

A single random sample of 16 different cereals is obtained, and the sugar content (in grams of sugar per gram of cereal) is measured for each cereal selected. Those amounts have a mean of 0.295 gram and a standard deviation of 0.168 gram. The amount of sugar in all cereals is assumed to be normally distributed.

Use a 0.05 significance level to test the claim of a cereal lobbyist that the mean of all cereals is less than 0.3 gram.

Hypotheses:                            [ Select ]                       ["Ho: μ = 0.3 vs. Ha: μ < 0.3", "Ho: μ = 0.3 vs. Ha: μ > 0.3", "Ho: μ < 0.3 vs. Ha: μ > 0.3", "Ho: μ < 0.3 vs. Ha: μ = 0.3"]      

Type of test:                            [ Select ]                       ["Left-tailed", "Right-tailed", "Two-tailed"]      

t statistic =                            [ Select ]                       ["-0.119", "0.119", "-1.753", "1.753"]         (calculated with formula)

Degrees of freedom =                            [ Select ]                       ["14", "15", "16", "17"]      

significance level =                            [ Select ]                       ["0.01", "0.05"]      

Critical t value =                            [ Select ]                       ["1.753", "-1.753", "2.131", "-2.131"]         (from table 10.1)

Decision:                            [ Select ]                       ["Reject the null hypothesis.", "Do not reject the null hypothesis."]      

Conclusion:                            [ Select ]                       ["We have sufficient evidence to support the claim that the mean for all cereals is less than 0.3 gram.", "We do not have sufficient evidence to support the claim that the mean for all cereals is less than 0.3 gram."]      

Answer 1

1)

sample mean, xbar = 0.295
sample standard deviation, s = 0.168
sample size, n = 16
degrees of freedom, df = n - 1 = 15

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.131

t value = 2.131

ME = tc * s/sqrt(n)
ME = 2.131 * 0.168/sqrt(16)
Margin of Error = 0.0895

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (0.295 - 2.131 * 0.168/sqrt(16) , 0.295 + 2.131 * 0.168/sqrt(16))
CI = (0.2055 , 0.3845)

2)

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 0.3
Alternative Hypothesis, Ha: μ < 0.3

left tailed test

Critical value of t is -1.753.

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (0.295 - 0.3)/(0.168/sqrt(16))
t = -0.119

Do not reject the null hypothesis."]

"We do not have sufficient evidence to support the claim that the mean for all cereals is less than 0.3 gram