Question

The economic dynamism, which is the index of productive growth in dollars for countries that are designated by the World Bank as middle-income are in table #1. Countries that are considered high-income have a mean economic dynamism of 60.29. Do the data show that the mean economic dynamism of middle-income countries is less than the mean for high-income countries? Test at the 5% level.

Table #1:

25.8057	37.4511	51.915	43.6952	47.8506	43.7178	58.0767
41.1648	38.0793	37.7251	39.6553	42.0265	48.6159	43.8555
49.1361	61.9281	41.9543	44.9346	46.0521	48.3652	43.6252
50.9866	59.1724	39.6282	33.6074	21.6643

a)      What is the appropriate test for this case?   

b)      What are the assumptions to run the test?   

c)      What is the null hypothesis?   

d)      What is the alternative hypothesis?   

e)      Determine if this test is left-tailed, right-tailed, or two-tailed.

f)       What is the significance level?

g)      What is the test statistics?   

h)      What is the p-value?

i)       Do we reject the null hypothesis? Why?   

j)       What is the conclusion?

k)      What is 95% confidence interval for the population mean?

l)       Interpret the confidence interval.

Include a cover sheet and the original statement in your solution.

You are asked to use an approved 12-point font, a professional layout and proper grammar.

Answer 1

The given same as

25.8057
41.1648
49.1361
50.9866
37.4511
38.0793
61.9281
59.1724
51.915
37.7251
41.9543
39.6282
43.6952
39.6553
44.9346
33.6074
47.8506
42.0265
46.0521
21.6643
43.7178
48.6159
48.3652
58.0767
43.8555
43.6252

According to question

a) Here we will apply t test as the sample size is less than 30.

b)Assumption used as :

The sample is randomly selected

The sample is approximately normal.

The sample is taken from Noemally distributed population.

c) The Null hypothesis

Ho: =60.29

d)The alternate hypothesis is

Ha : <60.29

e) Since the alternate hypothesis is Ha : M<60.29 we apply left tail test .

f) Here the significance level is 5 %(0.05)

g) The test statitistcs calculated as

The t-value is -9.228421

h)The p value computed as

The P-Value is < 0.00001

i) We reject null hypotheis becaluse the p value is less than level of significance. and it lies in area of rejection.

j) Since the p value is less than the level of significance hence we reject the null hypothesis and enough evidence to support the claim thet mean is less than 60.29.

k) Confidence interval

he formula for estimation is:

μ = M ± t(sM)

where:

M = sample mean
t = t statistic determined by confidence level
sM = standard error = √(s2/n)

Calculation

M = 43.87
t = 2.06
sM = √(9.072/26) = 1.78

μ = M ± t(sM)
μ = 43.87 ± 2.06*1.78
μ = 43.87 ± 3.6635

Result

95% CI [40.2065, 47.5335].

we can be 95% confident that the population mean (μ) falls between 40.2065 and 47.5335.