Find the circumference of the object. Use 3.14 or 22/7

A restaurant chain has two locations in a medium-sized town and, believing that it has oversaturated the market for its food, is

scoray [572]

Answer:

Yes, since the test statistic value is greater than the critical value.

Step-by-step explanation:

Data given and notation

$\bar X_{1}=360000$ represent the mean for the downtown restaurant

$\bar X_{2}=340000$ represent the mean for the freeway restaurant

$s_{1}=50000$ represent the sample standard deviation for downtown

$s_{2}=40000$ represent the sample standard deviation for the freeway

$n_{1}=36$ sample size for the downtown restaurant

$n_{2}=36$ sample size for the freeway restaurant

t would represent the statistic (variable of interest)

$\alpha=0.05$ significance level provided

Develop the null and alternative hypotheses for this study?

We need to conduct a hypothesis in order to check if the true mean of revenue for downtown is higher than for freeway restaurant, the system of hypothesis would be:

Null hypothesis: $\mu_{1}\leq \mu_{2}$

Alternative hypothesis: $\mu_{1} > \mu_{2}$

Since we don't know the population deviations for each group, for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Determine the critical value.

Based on the significance level $\alpha=0.05$ and $\alpha/2=0.025$ we can find the critical values from the t distribution dith degrees of freedom df=36+36-2=55+88-2=70, we are looking for values that accumulates 0.025 of the area on the right tail on the t distribution.

For this case the value is $t_{1-\alpha/2}=1.66$

Calculate the value of the test statistic for this hypothesis testing.

Since we have all the values we can replace in formula (1) like this:

$t=\frac{360000-340000}{\sqrt{\frac{50000^2}{36}+\frac{40000^2}{36}}}}=1.874$

What is the p-value for this hypothesis test?

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

$p_v =P(t_{70}>1.874)=0.033$

Based on the p-value, what is your conclusion?

Comparing the p value with the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the true mean for the downtown revenue restaurant seems higher than the true mean revenue for the freeway restaurant.

The best option would be:

Yes, since the test statistic value is greater than the critical value.

5 0

3 years ago

Using 50 boxes of nails a carpenter a

lapo4ka [179]

450=50x this should be correct :)

8 0

3 years ago

You need a loan for $1000, and you can take this loan out with Company A or Company B. Company A charges an 8% interest rate and

Wittaler [7]

Answer: Company b

Step-by-step explanation:

because company B is only 5percent it is better and can pay in 8 years

5 0

3 years ago

in speech class,you lose 3 points for every 30 seconds you go over the time limit.your speech is 90 seconds over the time limit

Cerrena [4.2K]

You’d loose 9 points if that’s the question

4 0

3 years ago

Devin is buying 2 concert tickets. The concert tickets have a regular price of $40 each. Devin has a coupon that gives a 5% disc

Svetach [21]

Given :

Devin is buying 2 concert tickets. The concert tickets have a regular price of $40 each.

Devin has a coupon that gives a 5% discount off of the regular price of the tickets. He will then pay a 10% purchasing fee on the discounted price of the tickets.

To Find :

Devin's total cost of the two tickets after the discount and the fee.

Solution :

Discount on price, D = 40×0.05 = $2 .

Price of product after discount, P = $( 40 - 2 ) = $38 .

Now, total price after adding purchasing fee is :

T = P + (P×0.10)

T = 38 + (38×0.10)

T = 38 + 3.8

T = $41.8

Therefore, Devin's total cost of the two tickets after the discount and the fee is $41.8 .

Hence, this is the required solution.

5 0

3 years ago