Mathematics, 12.02.2020 05:39, Stetson02

Hugger Polls contends that an agent conducts a mean of 53 in-depth home surveys every week. A streamlined survey form has been introduced, and Hugger wants to evaluate its effectiveness. The number of in-depth surveys conducted during a week by a random sample of agents are: <?xml:namespace prefix = o ns = "urn:schemas-microsoft-com:office:o ffice" /?>

53 57 50 55 58 54 60 52 59 62 60 60 51 59 56

At the .05 level of significance, can we conclude that the mean number of interviews conducted by the agents is more than 53 per week?

(a) What is the decision rule? (Round your answer to 3 decimal places.)

Reject H0 : ? ? 53 and accept H1 : ? > 53 when the test statistic is greater than

(b) The value of the test statistic is . (Round your answer to 2 decimal places.)

Answers: 1

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Answers

answered: cyaransteenberg

a) For this case we need to find first the degrees of freedom given by:

df= n-1= 15-1 =14

Now we need to find a quantile on the t distribution with 14 degrees of freedom that accumulates 0.05 of the area on the right and 0.95 on the left. And this value is: $t_{critc}=1.761$

And we can obtained with the following Excel code :"=T.INV(0.95,14)"

So for this case we will reject the null hypothesis if $t_{calculated}t_{critc}$ or if $t_{calculated}1.761$

b) $t=\frac{56.4-53}{\frac{3.74}{\sqrt{15}}}=3.52$

Step-by-step explanation:

Data given and notation

We have the following dataset:

53 57 50 55 58 54 60 52 59 62 60 60 51 59 56

We can calculate the mean and deviation with the following formulas:

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

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

And the results are:

$\bar X=56.4$ represent the sample mean

s=3.74 represent the sample standard deviation

n=15 sample size

$\mu_o =53$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

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

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 53, the system of hypothesis would be:

Null hypothesis: $\mu \leq 53$

Alternative hypothesis: $\mu 53$

If we analyze the size for the sample is > 30 but we don't know the population deviation so 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".

a) Decision rule

For this case we need to find first the degrees of freedom given by:

df= n-1= 15-1 =14

Now we need to find a quantile on the t distribution with 14 degrees of freedom that accumulates 0.05 of the area on the right and 0.95 on the left. And this value is: $t_{critc}=1.761$

And we can obtained with the following Excel code :"=T.INV(0.95,14)"

So for this case we will reject the null hypothesis if $t_{calculated}t_{critc}$ or if $t_{calculated}1.71$

b) Calculate the statistic

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

$t=\frac{56.4-53}{\frac{3.74}{\sqrt{15}}}=3.52$

Conclusion

As we can see our calculated value is higher than the critical value 3.52>1.76 so then we are on the rejection zone and we can conclide that that we have enough evidence to reject the null hypothesis, so the means seems to be higher than 53 at 5% of signficance.

answered: Guest

The answer of the above is

answered: Guest

4 candies

step-by-step explanation:

7 x 28 = 196

200 - 196 = 4 candies remaining.