A random sample of 10 college students was drawn from a large university. Their ages are 22, 17, 27, 20, 23, 19, 24, 18, 19, and 24 years. Test the hypothesis to determine if we can infer at the 5% significance level that the population mean is less than 20.

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belgrad belgrad · Answer 1 · 2020-03-29T17:52:19+02:00

Answer:

The test statistic t = 1.219 < 2.262 at 5% level of significance

we accept significance of level that the population mean is less than 20.

Step-by-step explanation:

Given ages are 22, 17, 27, 20, 23, 19, 24, 18, 19, and 24

The random sample (n) = 10

Null hypothesis (H0): μ = 20

Alternative hypothesis(H1) : μ < 20 (left tailed test)

we will use statistic 't' distribution with small sample 10 < 30

[tex]t = \frac{χ-μ}{\frac{S.D}{\sqrt{n-1} } }[/tex]

mean (χ) = sum of observations divided by no of observations

mean(x) = ∑x / n = [tex]\frac{22+17+27+20+23+19+24+18+19+24}{10} = 21.3[/tex]

x x - mean (x-mean)^2

22 22-21.3 = 0.7 0.49

17 17-21.3 = -4.3 18.49

27 27-21.3 = 5.7 32.49

20 20-21.3 =-1.3 1.69

23 23-21.3 = 1.7 2.89

19 19-21.3 = -2.3 5.29

24 24-21.3 = 2.7 7.29

18 18-21.3 = -3.3 10.89

19 19-21.3 = -2.3 5.29

24 24-21.3 = -=2.7 7.29

[tex]S^{2} = \frac{∑(x-mean)^2}{n-1}= \frac{92.1}{10-1}[/tex]

[tex]S^{2} = \frac{92.1}{9} = 10.2[/tex]

S = 3.198

The test statistic (t) = [tex]t = \frac{21.3-20}{\frac{3.198}{\sqrt{10-1} } }[/tex]

t = 1.219

The degrees of freedom = n-1 = 10-1 =9

Tabulated value of t for 9 degrees of freedom at 5% level of significance

= 2.262

since calculated t < tabulated t we accept the null hypothesis

we accept significance of level that the population mean is less than 20.