What is the number in front of the variable<br>

1. The U.S. Energy Information Administration claimed that U.S. residential customers used an average of 10,290 kilowatt hours (

vladimir2022 [97]

Answer:

Kindly check explanation for (1)

2.)

B. We fail to reject the null hypothesis and conclude that there is insufficient evidence at a 0.02 level of significance to support the power company claim that the mean amount of electricity for their residents is more than the national average.

Step-by-step explanation:

1.)

H0 : μ = 10290

H0 : μ > 10290

n = 153, sample mean, xbar = 10586 ; population standard deviation, σ = 2509 ; α = 0.05

Test statistic:

(xbar - μ) ÷ (σ/√(n))

(10586 - 10290) ÷ (2509/√(153))

296 / 202.84062

Test statistic (Z) = 1.46

We obtain the Pvalue :

Decision Region :

Reject H0 ; if Pvalue < α

Pvalue from Z test statistic calculator :

Pvalue = 0.072

Pvalue > α ; Hence, we fail to reject the Null ; there isn't sufficient evidence to support the power company's claim

2.)

H0 : μ = 10290

H0 : μ > 10290

n = 153, sample mean, xbar = 10446 ; population standard deviation, σ = 2509 ; α = 0.02

Test statistic:

(xbar - μ) ÷ (σ/√(n))

(10446 - 10290) ÷ (2509/√(153))

156 / 202.84062

Test statistic (Z) = 0.769

We obtain the Pvalue :

Decision Region :

Reject H0 ; if Pvalue < α

Pvalue from Z test statistic calculator :

Pvalue = 0.221

Pvalue > α ; Hence, we fail to reject the Null ;

B. We fail to reject the null hypothesis and conclude that there is insufficient evidence at a 0.02 level of significance to support the power company claim that the mean amount of electricity for their residents is more than the national average.

5 0

3 years ago

In a survey of a group of men, the heights in the 20-29 age group were normally distributed, with a mean of 67 inches and a s

liq [111]

Answer:

a) 20.33% probability that the participant is less than 64.5 inches.

b) 33.72% probability that the participant is more than 68.25 inches

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 67, \sigma = 3$

a.) Find the probability that the participant is less than 64.5 inches?

This is the pvalue of Z when X = 64.5.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{64.5 - 67}{3}$

$Z = -0.83$

$Z = -0.83$ has a pvalue of 0.2033

20.33% probability that the participant is less than 64.5 inches.

b.) Find the probability that the participant is more than 68.25 inches?

This is 1 subtracted by the pvalue of Z when X = 68.25.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{68.25 - 67}{3}$

$Z = 0.42$

$Z = 0.42$ has a pvalue of 0.6628

1 - 0.6628 = 0.3372

33.72% probability that the participant is more than 68.25 inches

3 0

3 years ago

Hey guys! I'm really stuck on numbers 2 and 3. Can you help?

Molodets [167]

The second one is probably 70 and the third one is 21

4 0

3 years ago

A table<br> cloth is 82 inches<br> long. What is its length in<br> feet and inches?

Ratling [72]

The answer is 6 feet 10 inches

4 0

2 years ago

Each of forty-six pet owners was asked, "Does your pet eat more wet food or dry food?" Here are the results. 9 cat owners and 13

Ahat [919]

Answer:

Cat owner Dog owner Total

Wet food 9 13 22

Dry food 10 14 24

Total 19 27 46

Step-by-step explanation:

It is given that 9 cat owners and 13 dog owners chose "Wet food". 10 cat owners and 14 dog owners chose "Dry food".

Cat owners-Wet food = 9

Dog owners-Wet food =13

Cat owners-Dry food = 10

Dog owners-Dry food =14

We need to construct a two-way frequency table for the data.

Take owners in the first row and type of food in first column and construct a two-way frequency as shown below:

Cat owner Dog owner Total

Wet food 9 13 22

Dry food 10 14 24

Total 19 27 46

3 0

3 years ago

What is the number in front of the variable​

What is the number in front of the variable