2. A sample of cereal boxes is randomly selected and the sugar contents (grams o
ID: 3053840 • Letter: 2
Question
2. A sample of cereal boxes is randomly selected and the sugar contents (grams of sugar per gram of cereal) are recorded. Those amounts are summarized with these statistics: n 40,-0.295 g, s 0.168 g. A cereal lobbyist claims that the mean for all cereals is less than 0.3 g. Construct a 95% confidence interval for the mean sugar content of all cereals (H) and provide an interpretation for the interval Test whether the cereal lobbyist's claim about the mean sugar content of all cereals at the 0.05 significance level. Provide the null and alternative hypotheses, the test statistic, degrees of freedom, p-value, result (reject or fail to reject Ho), and conclusions. Explain how you could have used the confidence interval constructed in part (a) to determine whether to reject or fail to reject null hypothesis for the test in part (b). Be specific Explain how you could have determined whether to reject or fail to reject the null hypothesis for the test in part (b) using the test statistic and critical value without calculating the p-value. Be specific. a. b. c. d.Explanation / Answer
a)
CI for 95%
n = 40
mean = 0.295
z-value of 95% CI = 1.9600
std. dev. = 0.168
SE = std.dev./sqrt(n) = 0.02656
ME = z*SE = 0.05206
Lower Limit = Mean - ME = 0.24294
Upper Limit = Mean + ME = 0.34706
95% CI (0.2429 , 0.3471 )
As sample size is greater than 30, we use z-value table.
b)
Below are the null and alternate hypothesis
H0: mu >= 0.3
Ha: mu < 0.3
Test statistic, z = (0.295 - 0.3)/(0.168/sqrt(40)) = -0.1882
p-value = 0.4253
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 mean is less than 0.3g
c)
As the hypothesised value 0.3 lies within the confidence interval, we fail to reject the null hypothesis.
d)
Critical value for 0.05 significance level is -1.65 (value is negative as this is left tailed test)
As test statistic z = -0.1882 is greater than the critical value, we fail to reject the null hypothesis.
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