Double s, then multiply t by the result

Fowle Marketing Research Inc., bases charges to a client on the assumption that telephone surveys can be completed in a mean tim

sergejj [24]

Answer:

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.

7 0

3 years ago

A. 7 ft B. 8 ft C. 49 ft D. 576 ft

AleksAgata [21]

Your equation is gonna be a^2-24^2=25^2 and your answer is 7.

4 0

2 years ago

Read 2 more answers

500 cats were chosen. 52% of the cats liked to sleep inside the house. Fish was the favorite dish of 25% of those cats, while 62

Alex777 [14]

You would need to take 500 * .52 = 260 next take 260 * .25 to find how many cats like fish which equals 65, the probability that it likes fish and it sleeps inside can be found by taking .52 * .25 = .13 so there would be a 13% chance of picking a cat that sleeps indoors and likes fish, 435 cats dont like fish found by taking 500-65=435 and finally to find how many cats sleep outside take 500-260=240 then take 240 * .625 = 150 then 240-150 = 90 so 90 cats like to sleep outside and like fish thats it!=) Hope this helps!

3 0

3 years ago

Read 2 more answers

Which is true 10.59>10.60 10.590=10.59 10.43<10.24 9.0005=9.0050

mart [117]

I think the second i didnt make it on paper but its the one that more convence me

4 0

3 years ago

It is 75 miles from my house to Georgia aquarium on the map it measures one 3/4 inches what is the scale of the map

harkovskaia [24]

Need more to answer this

3 0

3 years ago