It is believed that the average amount of money spent per U. S. household per week on food is about $99, with standard deviation $10. A random sample of 25 households in a certain affluent community yields a mean weekly food budget of $100. We want to test the hypothesis that the mean weekly food budget for all households in this community is higher than the national average. State the null and alternative hypotheses for this test, the test statistic and determine if the results significant at the 5% level.

H0 : μ = 99

H1 : μ > 99

Z statistic = 0.5

Result is not significant at 5% level ;

Step-by-step explanation:

Given that :

σ = 10

μ = 99

Sample size, n = 25

H0 : μ = 99

H1 : μ > 99

The test statistic :

(xbar - μ) ÷ (σ/√(n))

Since the population standard deviation is known, we use the Z statistics.

Z = (100 - 99) ÷ (10/√(25))

Z = 1 / 2

Z = 0.5

The Z critical value at α = 0.05

Zcritical = 1.645

Zstatistic < Zcritical ; Hence, we fail to reject the H0.

i did it myself and i got 4.2 (b) hope its right

step-by-step explanation:

1. x> -5

2. x< 8

step-by-step explanation:

1. 2x> -10

divide by 2 from both sides of equation.

2x/2> -10/2

simplify, to find the answer.

-10/2=-5

x> -5 is the correct answer.

2. 3x< 24

divide by 3 from both sides of equation.

3x/3< 24/3

simplify, to find the answer.

24/3=8

x< 8 is the correct answer.

i hope this you, and have a wonderful day!