Answer to Question #175137 in Statistics and Probability for jona villanueva

Question

According to a study done last year, the average monthly expenses for mobile phone loads of college students in San Mateo, Rizal was P 400.00. A statistics student believes that this amount has decreased since January of this year. Is there a reason to believe that this amount has decreased if a random sample of 50 students has an average monthly expense for mobile phone loads of P 380.00? Use .05 level of significance. Assume the population standard deviation is P 75.00.  

Step 1: State the null and alternative hypotheses.

(1.) Null Hypothesis (Ho):

(2.) Alternative Hypothesis(Ha):

(3.) Write Ho in words:

(4.) Write Ha in words:

Step 2: Level of significance and the Type of test.

(5.) Level of Significance: α =

(6.) Two-tailed Test or One-tailed Test

Step 3: Compute the test statistic.

(7.) = (8.)  = ______ (9.) ẟ = ______ (10.) n = _or   s =

ComputationEncircle the formula to be used

4: Determine the critical value.

Answer 1

We have that:

$\mu=400$

$\sigma=75$

$n=50$

$\bar x=380$

$\alpha=0.05$

$H_0:\mu=400$

$H_a:\mu<400$

H₀ : "the average monthly expenses for mobile phone loads of college students in San Mateo, Rizal is P 400.00"

H_a : "the average monthly expenses for mobile phone loads decreased"

The hypothesis test is left-tailed.

The population standard deviation is known and the sample size is large (n>30) so we use z-test.

The critical value for α = 0.05 is Z_0.05 = –1.64

The critical region is z < –1.645.

Test statistic:

$$Z_{test}=\frac{\bar x - \mu}{\frac{\sigma}{\sqrt n}}=\frac{380- 400}{\frac{75}{\sqrt 50}}=-1.89$$

Since –1.89 < –1.645 thus the Z_test falls in the rejection region we reject the null hypothesis.

At the 5% significance level the data do provide sufficient evidence to support the claim. We are 95% confident to conclude that the average monthly expenses for mobile phone loads decreased.