Your answer is partially correct. Try again. M&M; Consumption A study investigat

Question

Your answer is partially correct. Try again. M&M; Consumption A study investigated how many M&Ms; were taken by a sample of 20 college students who were presented with candies in a 24 -ounce bowl. Compare the numbers of M&Ms; taken by males and females to investigate whether one gender tends to take more candies than the other. dy 41 61 44 111 21 42 7 11 36 62 92 19 35 50 104 97 43 62 33 32 Males Females a. Produce a 90% confidence interval for remale males. Round your answers to two decimal places. ?male, the difference in population means between females and 36.66 The confidence interval is-10.44 b. Interpret what this interval reveals. Enter your answers in increasing order, rounded to two decimal places : You are 90% confident the mean number of M&Ms; taken by females from such a bowl is between -10.44 and 36.66 more than the mean number of M&Ms; taken by males c. A two-sample -test was conducted to test whether males and females differ with regard to how many M&Ms; they take from a 24 -ounce bowl. At the o-0.10 significance level, Ho was not rejected. Are the test decision and the confidence interval consistent with each other? Click if you would like to Show Work for this question: Open Show Work

Explanation / Answer

Using R codes,

CODE:

males <- c(41,61,44,111,21,42,7,11,36,62)
females <- c(92,19,35,50,104,97,43,62,33,32)
t.test(females,males,conf.level = 0.9)

OUTPUT:

> males <- c(41,61,44,111,21,42,7,11,36,62)
> females <- c(92,19,35,50,104,97,43,62,33,32)

> t.test(females,males,conf.level = 0.9)

Welch Two Sample t-test

data: females and males

t = 0.96468, df = 17.995, p-value

= 0.3475

alternative hypothesis: true difference in means is not equal to 0

90 percent confidence interval:

-10.44822 36.64822

sample estimates:

mean of x mean of y

56.7 43.6

So, here the 90% confidence interval is (-10.45, 36.65).

Here P- value is greater than 0.01. So, we can't reject the null hypothesis,

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