Answer:
Correct option: (B)
Step-by-step explanation:
A one-sample <em>t</em>-test can be performed to determine whether the mean weight of female college students is still 59.4 kg.
The hypothesis can be defined as:
<em>H₀</em>: The mean weight of female college students has changed.
<em>Hₐ</em>: The mean weight of female college students has not changed.
The information provided is:

As the there is no information about the population standard deviation we will use a <em>t</em>-test for the mean.
The test statistic is:

Decision rule:
If the if the <em>p</em>-value of the test is less than the significance level of the test <em>α</em> then the null hypothesis will be rejected and vice versa.
The degrees of freedom of the test is:

The test is two-tailed.
Compute the <em>p</em>-value of the test as follows:

*Use a <em>t</em>-table for the probability.
The <em>p</em>-value = 0.13 > <em>α</em> = 0.10
The null hypothesis was failed to be rejected.
As the null hypothesis was rejected it can be concluded that there is not sufficient evidence that the mean weight of female students has changed.
Answer:
95 ≤ 8 < 1/7n
Step-by-step explanation:
95 ≤ 8 < 1/7n
no more = less than or equal to
product of a number n and 1/7 = 1/7•n = 1/7n
Answer:
i) 0.872
ii) 0.300
iii) 0.76
iv) 0.704
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = $261.50 per month
Standard Deviation, σ = $16.25
We are given that the distribution of monthly food cost for a 14- to 18-year-old male is a bell shaped distribution that is a normal distribution.
Formula:
a) P(Less than $280)
Calculation the value from standard normal z table, we have,
b) P(More than $270)
P(x > 270)
Calculation the value from standard normal z table, we have,
c) P(More than $250)
P(x > 250)
Calculation the value from standard normal z table, we have,
d) P(Between $240 and $275)

e) Thus, 0.704 is the probability that the monthly food cost for a randomly selected 14- to 18-year-old male is between $240 and $275.