Answer to Question #188411 in Statistics and Probability for sean

Question

a. state the null and alternative hypothesis

b. compute the test statistics

c. determine the critical value and the rejection region

d. draw a conclusion

1. It is claimed that the average weight of babies at birth is 3.4 kg. The average weight of a random sample of 30 newly born babies was determined. It was found out that the average weight was 3.1 kg.

Is there a reason to believe that the average weight of the babies at birth is not 3.4 kg? Assume that the population standard deviation is 1.1 kg. (Use 0.05 level of significance.)

2. Last year, a real estate agent earned an average of ₱40,250.00 a month. Suppose you recently selected a random sample of 25 real estate agents and you have determined how much each of them earns each month. Your computations of their earnings resulted to an average of ₱40,400.00 with a standard deviation of ₱225.00. Using a 0.01 level of significance, can it be concluded that the average monthly earning of real estate agents has increased? Assume normality in the population

Answer 1

1.

(a) Ho: µ₁ = µ₀, (the average weight of babies is not different from 3.4kgs (µ₀))

Ha: µ₁ $\neq$ µ₀, (µ₀ =3.4kg), (the average weight of babies is different from 3.4kgs)

Level of Significance: α=0.05

(B) Test- statistic: Z- statistics (this is because the sample size is large enough, $n\geq 30$ ). Thus we have $Z=\frac{\bar{X}-\mu_0}{\frac{s}{\sqrt{n}}}$

Tails in Distribution: Two-tailed

(C) Reject H₀ if $Z\geq 1.960 \text{ or if } Z\leq -1.960$.

Test statistics

$Z=$ $\frac{\bar{X}-\mu_0}{\frac{s}{\sqrt{n}}}=\frac{3.1-3.4}{\frac{1.1}{\sqrt{30}}}=-1.494$

We fail to reject H₀ because $Z=-1.494>-1.960$.

(D) We do not have statistically significant evidence at α=0.05, to show that the average weight of babies at birth is not 3.4kg.

2.(a)$H_o: µ_1 = µ_0$ , the average monthly earning of real estate agents has increased

   $H_a: µ_1 \neq µ_0,$

Level of Significance: α=0.01

(B) Test- statistic: Z- statistics (this is because the sample size is large enough, n\le 30 ). Thus we

have $t=\frac{\bar{X}-\mu_0}{\frac{s}{\sqrt{n}}}$

$t=\dfrac{40250-40400}{\frac{225}{\sqrt{25}}}$

$=\dfrac{-150\times 5}{225}=-3.33$

Tails in Distribution: single -tailed

(C) Reject $H0 \text{ if } t\le -3.33$

Test statistics

$t= \frac{\bar{X}-\mu_0}{\frac{s}{\sqrt{n}}}=\frac{40250-40400}{\frac{225}{\sqrt{25}}}=-3.33$

The value of t at $\alpha=0.01$ , and degree of freedom $n-1 i.e 24 \text{ is }2.492.$

(D) We do not have statistically significant evidence at α=0.01,the average monthly earning of real estate agents has not increased.