In 1960, census results indicated that the age at which men in a certain region first married had a mean of 23.5 years. We want to find out if the mean age of first marriage has changed/differed from 23.5 years (µ≠23.5) since then. The 40 men in our sample first married at an average age of 24.3 years, with a sample standard deviation s of 5.3 years. The P-value is 0.346. State the conclusion using α = 0.10.

There is not sufficient evidence that the mean age of first marriage differs the mean age in 1960.

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 23.5 years

Sample mean, = 24.3 years

Sample size, n = 40

Alpha, α = 0.10

Sample standard deviation, s = 5.3 years

P-value = 0.346

First, we design the null and the alternate hypothesis

We use Two-tailed t test to perform this hypothesis.

Formula:

Putting all the values, we have

P-value = 0.346

Since the calculated p-value is greater than the significance level, we fail to reject the null hypothesis and accept the null hypothesis.

Conclusion:

Thus, there is not sufficient evidence that the mean age of first marriage differs the mean age in 1960.

simplifying

5(x + -6) = 2(x + 3)

reorder the terms:

5(-6 + x) = 2(x + 3)

(-6 * 5 + x * 5) = 2(x + 3)

(-30 + 5x) = 2(x + 3)

reorder the terms:

-30 + 5x = 2(3 + x)

-30 + 5x = (3 * 2 + x * 2)

-30 + 5x = (6 + 2x)

solving

-30 + 5x = 6 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-30 + 5x + -2x = 6 + 2x + -2x

combine like terms: 5x + -2x = 3x

-30 + 3x = 6 + 2x + -2x

combine like terms: 2x + -2x = 0

-30 + 3x = 6 + 0

-30 + 3x = 6

add '30' to each side of the equation.

-30 + 30 + 3x = 6 + 30

combine like terms: -30 + 30 = 0

0 + 3x = 6 + 30

3x = 6 + 30

combine like terms: 6 + 30 = 36

3x = 36

divide each side by '3'.

x = 12

simplifying

x = 12

this should .