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igomit [66]
3 years ago
6

I need help and fast

Mathematics
1 answer:
suter [353]3 years ago
4 0
Why are you taking a map test in january
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A survey collected data on annual credit card charges in seven different categories of expenditures: transportation, groceries,
Andru [333]

Answer:

A. Null and alternative hypothesis:

H_0: \mu_d=0\\\\H_a:\mu_d\neq 0

B. Yes. At a significance level of 0.05, there is enough evidence to support the claim that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

P-value = 0.00002

C. As the difference is calculated as (population 1 − population 2), being population 1: groceries and population 2: dinning out, and knowing there is evidence that the true mean difference is positive, we can say that the groceries annual credit card charge is higher than dinning out annual credit card charge.

The point estimate is the sample mean difference d=$840.

The 95% confidence interval for the mean difference between the population means is (490, 1190).

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

Then, the null and alternative hypothesis are:

H_0: \mu_d=0\\\\H_a:\mu_d\neq 0

The significance level is 0.05.

The sample has a size n=42.

The sample mean is M=840.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=1123.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{1123}{\sqrt{42}}=173.2827

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{840-0}{173.2827}=\dfrac{840}{173.2827}=4.848

The degrees of freedom for this sample size are:

df=n-1=42-1=41

This test is a two-tailed test, with 41 degrees of freedom and t=4.848, so the P-value for this test is calculated as (using a t-table):

\text{P-value}=2\cdot P(t>4.848)=0.00002

As the P-value (0.00002) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

We have to calculate a 95% confidence interval for the mean difference.

The t-value for a 95% confidence interval and 41 degrees of freedom is t=2.02.

The margin of error (MOE) can be calculated as:

MOE=t\cdot s_M=2.02 \cdot 173.283=350

Then, the lower and upper bounds of the confidence interval are:

LL=M-t \cdot s_M = 840-350=490\\\\UL=M+t \cdot s_M = 840+350=1190

The 95% confidence interval for the mean difference is (490, 1190).

7 0
3 years ago
Am gonna give branliest!!!
blagie [28]
The answer is B. -12°f
6 0
3 years ago
Read 2 more answers
Suppose △LMN≅△PQR .
Tanya [424]
First four answers are correct as congruent parts of congruent triangles are congruent but L Is not the opposite to M so they wouldn’t be congruent nor does NL or PQ have the same connection
6 0
3 years ago
yo plz help this is my fourth time posting this question plz i need this plz plz plz plz plz PLEASE. IT IS DRAINING MY POINTS :(
timama [110]

Answer:

A

Step-by-step explanation:

7 0
3 years ago
Two batches of soup will serve 8 people. Lonnie needs to make enough soup for 120 people. If the recipe calls for 2 onions for o
ELEN [110]
I am not entirely sure but I think the answer would be 60

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3 years ago
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