Divide 15 into 372 and what is the answer
2 answers:
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Answer:
(a) Decision rule for 0.01 significance level is that we will reject our null hypothesis if the test statistics does not lie between t = -2.651 and t = 2.651.
(b) The value of t test statistics is 1.890.
(c) We conclude that there is no difference in the mean number of times men and women order take-out dinners in a month.
(d) P-value of the test statistics is 0.0662.
Step-by-step explanation:
We are given that a recent study focused on the number of times men and women who live alone buy take-out dinner in a month.
Also, following information is given below;
Statistic : Men Women
The sample mean : 24.51 22.69
Sample standard deviation : 4.48 3.86
Sample size : 35 40
<em>Let </em><em> = mean number of times men order take-out dinners in a month.</em>
<em /><em> = mean number of times women order take-out dinners in a month</em>
(a) So, Null Hypothesis, :
= 0 {means that there is no difference in the mean number of times men and women order take-out dinners in a month}
Alternate Hypothesis, :
0 {means that there is difference in the mean number of times men and women order take-out dinners in a month}
The test statistics that would be used here <u>Two-sample t test statistics</u> as we don't know about the population standard deviation;
T.S. = ~
where, = sample mean for men = 24.51
= sample mean for women = 22.69
= sample standard deviation for men = 4.48
= sample standard deviation for women = 3.86
= sample of men = 35
= sample of women = 40
Also, =
= 4.16
So, <u>test statistics</u> = ~
= 1.890
(b) The value of t test statistics is 1.890.
(c) Now, at 0.01 significance level the t table gives critical values of -2.651 and 2.651 at 73 degree of freedom for two-tailed test.
Since our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u>we fail to reject our null hypothesis</u>.
Therefore, we conclude that there is no difference in the mean number of times men and women order take-out dinners in a month.
(d) Now, the P-value of the test statistics is given by;
P-value = P( > 1.89) = 0.0331
So, P-value for two tailed test is = 2 0.0331 = <u>0.0662</u>
Answer:
$2.79 per pound of apples.
0.36 pounds per dollar.
The unit rate representing cost of per pound apples is typically used.
Step-by-step explanation:
We have been given that the weekly ad for a local grocery store advertises a 5 pound bag of organic apples for $13.95.
Since we know when rates are expressed as a quantity of 1, such as 100 meter per second or 5 miles per hour, they are called unit rates. We need to have 1 in our denominator to express two quantities as unit rate.
Let us find the unit rate in terms of cost of per pound bag of apples.
Therefore, our unit rate will be $2.79 per pound of apples.
Let us find unit rate of pounds of apples per dollar.
Therefore, our another unit rate will be 0.36 pounds per dollar.
Since, we purchase apples or other items according to their price per piece or their price per pound in our daily life, therefore, the unit rate representing cost of per pound apples is typically used.