Mathematics, 21.04.2020 22:47, oliviagilpin8p6lk1i

Fowle Marketing Research Inc., bases charges to a client on the assumption that telephone surveys can be completed in a mean time of 15 minutes or less. If a longer mean survey time is necessary, a premium rate is charged. A sample of 35 surveys provided the following survey times in minutes: 17;11;12;23;20;23;15;16;23;22;18;23 ;25;14;12;12;20;18;12;19;11;11;20;2 1;11;18;14;13;13;19; 16;10;22;18;23. Based upon past studies, the population standard deviation is assumed known with s = 4 minutes. Is the premium rate justified?

a. Formulate the null and alternative hypotheses for this application.
b. Compute the value of the test statistic.
c. What is the p-value?
d. At a = .01, what is your conclusion?

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answered: icantspeakengles

a) Null hypothesis: $\mu \leq 15$

Alternative hypothesis: $\mu 15$

b) $t=\frac{17-15}{\frac{4}{\sqrt{35}}}=2.958$

c) $p_v =P(t_{34}2.958)=0.0028$

d) If we compare the p value and the significance level given $\alpha=0.01$ we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 15 at 1% of signficance.

Step-by-step explanation:

Data given and notation

The sample mean is given by:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And the sample deviation is:

$s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X=17$ represent the sample mean

s=4 represent the sample standard deviation

n=35 sample size

$\mu_o =15$ represent the value that we want to test

$\alpha=0.01$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

p_v represent the p value for the test (variable of interest)

Part a State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is less or equal than 15, the system of hypothesis would be:

Null hypothesis: $\mu \leq 15$

Alternative hypothesis: $\mu 15$

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Part b Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{17-15}{\frac{4}{\sqrt{35}}}=2.958$

Part c P-value

The degrees of freedom are given by:

df = n-1= 35-1=34

Since is a right tailed test the p value would be:

$p_v =P(t_{34}2.958)=0.0028$

Part d Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 15 at 1% of signficance.

answered: Guest

if gloria has "carry over" minutes, (that is, any minutes paid for but not used in a month can be applied to her payment for minutes in the next month,) her total usage for the three months is: total minutes = 320 + 243 + 489 = 1052 minutes number of 200 minute blocks = 1052/200 = 5 r52 (that is 5 with a remainder of 52) rounding up the number of 200 minute blocks gives her usage for the three months as 6 blocks. the cost for six blocks is 6 x $25 = $150 she also has 200 - 52 or 148 carry over minutes if gloria pays each month and unused minutes are lost, applying the rounding up procedure to each month gives: first month usage of 320 minutes rounds up to 2 blocks of 200 minutes for a payment of 2 x $25 = $50 second month usage of 243 minutes rounds up to 2 blocks of 200 minutes for a payment of 2 x $25 = $50 third month usage of 489 minutes rounds up to 3 blocks of 200 minutes for a payment of 3 x $25 = $75 total payments for the three months are $50 + $50 + $75 = $175 get the plan with carry over minutes!

answered: Guest

answer: list the statements and i will follow up!

step-by-step explanation: