Question

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?

Answer 1

Answer:

(a) The median is 1478

(b) Null hypothesis H₀ : μ₁ = μ₂

Alternative hypothesis Hₐ : μ₁ > μ₂

(c) The test statistic  $t_{\alpha/2}$ 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,  $\bar{x}_{1}$= 1541.9

Standard deviation, σ₁ = 41.38224257

n₁ = 10

Below the median  

1471 1458 1456 1458 1453 1438 1428 1405 1395 1377

The mean,  $\bar{x}_{2}}$= 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;

$t_{\alpha/2} =\frac{(\bar{x}_{1}-\bar{x}_{2})}{\sqrt{\frac{\sigma_{1}^{2} }{n_{1}}-\frac{\sigma _{2}^{2}}{n_{2}}}}$

df = 10 - 1 = 9, α = 0.05

Therefore, the test statistic  $t_{\alpha/2}$ = 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.