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morpeh [17]
2 years ago
7

A random sample of n = 16 professors from a university has been selected; salaries have been plotted on the following Q-Q plot.

qqplot If we created a 95% confidence interval for salaries to be ($99,881, $171,172), how would we interpret that interval? Since n = 16 > 15, we can use the CLT to say we are 95% sure that all professors' salaries at this university are between $99,881 and $171,172. Since n = 16 > 15, we can use the CLT to say we are 95% sure the average of all professors' salaries at this university is between $99,881 and $171,172. We actually can't be 95% sure the average professor salary is in the interval, since the salaries are right-skewed and n = 16 < 30.
Mathematics
1 answer:
Semenov [28]2 years ago
5 0

Answer:

The objective of the confidence interval is to give a range in which the real mean of the population is placed, with a degree of confidence given by the level of significance.

The conclusion we can make is that there is 95% of probability that the mean of the population (professor's average salary) is within $99,881 and $171,172.

Step-by-step explanation:

This is a case in which, from a sample os size n=16, a confidence interval is constructed.

The objective of the confidence interval is to give a range in which the real mean of the population is placed, with a degree of confidence given by the level of significance. In this case, the probability that the real mean is within the interval is 95%.

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Hatshy [7]

Answer:

$2 for 1 rose bush and $2 for one shrub

Step-by-step explanation:

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32=2(2)+14(2)

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32=32

26=11(2)+2(2)

26=22+4

26=26

Since both equations come out correct the cost of 1 rose bush is 2 dollars and the cost of one shrub is 2 dollars. kinda cool

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3 years ago
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The average expenditure on Valentine's Day was expected to be $100.89 (USA Today, February 13, 2006). Do male and female consume
AfilCa [17]

Answer:

Yes. The male and female consumers differ in the amounts they spend.

Step-by-step explanation:

We can express the null and alternative hypothesis as:

H_0: \mu_m=\mu_w\\\\H_1:  \mu_m\neq\mu_w

It is assumed a significance level of 0.05.

The standard deviation of the difference of means is calculated as:

s=\sqrt{\frac{s_m^2}{n_m} +\frac{s_w^2}{n_w} } =\sqrt{\frac{35^2}{40} +\frac{20^2}{30} } =\sqrt{30.625+13.333} =\sqrt{43.958} =6.63

The test statistic is

t=\frac{(M_m-M_w)-0}{s}=\frac{135.67-68.64}{6.63}=10.11

The degrees of freedom are:

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The P-value for t=10.11 is P=0, so it is smaller than the significance level. The null hypothesis is rejected.

We can conclude that male and female consumers differ in the amounts they spend.

5 0
3 years ago
Harper knows he is 50 yards from school. The map on his phone shows that the school is 3/4 inch from his current location. How f
yulyashka [42]

Answer:

Harper is 200 yards away from her home if the map shows 3 inches

Step-by-step explanation:

50 yards = 3/4 inch

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I hope this helped and if it did I would appreciate it if you marked me Brainliest. Thank you and have a nice day!

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