Media experts claim that daily print newspapers are declining because of Internet access. Listed below, from left to right and then top to bottom, are the numbers of daily print newspapers in a large region for a recent sequence of years. First find the median, then test for randomness of the numbers above and below the median using alphaequals0.05. What do the results suggest?

(a) The median is 1478

(b) Null hypothesis H₀ : μ₁ = μ₂

Alternative hypothesis Hₐ : μ₁ > μ₂

(c) The test statistic   is 6.678155

(d) The p value is  1.2×10⁻¹¹

(e) We reject our null hypothesis H₀. In terms of the test, there is sufficient evidence to suggest that there is a trend in the numbers of daily newspaper because the probability for a decrease in the numbers is very high

Step-by-step explanation:

(a) Here we have the data as follows;

1623 1586 1574 1554 1544 1531 1519 1513 1491 1484 1478 1471 1458 1456 1458 1453 1438 1428 1405 1395 1377

The median is = 1478

Therefore we have above the median  

1623 1586 1574 1554 1544 1531 1519 1513 1491 1484

The mean,  = 1541.9

Standard deviation, σ₁ = 41.38224257

n₁ = 10

Below the median  

1471 1458 1456 1458 1453 1438 1428 1405 1395 1377

The mean,  = 1433.9

Standard deviation, σ₂ = 30.04812806

n₂ = 10

(b) Null hypothesis H₀ : μ₁ = μ₂

Alternative hypothesis Hₐ : μ₁ > μ₂

(c) The formula for t test is given by;

df = 10 - 1 = 9, α = 0.05

Therefore, the test statistic   = 6.678155

(d) The p value from statistical relations is Probability p = 1.2×10⁻¹¹

Critical z at 5% confidence level = 1.645

Since P << 0.05, e reject the

(e) Therefore, we reject our null hypothesis H₀. In terms of the test, there is sufficient evidence to suggest that there is a trend in the numbers of daily newspaper because the probability for a decrease in the numbers is very high.

step-by-step explanation:

first you will have to convert 3/4 into a decimal

which is .75

then you will have to divide 1.5 by .75

which is 2

no you have to divide