There are 25 stocks and this is their descriptive statistics. mean: 34.841 St De

Question

There are 25 stocks and this is their descriptive statistics.

mean: 34.841

St Dev: 38.36951

Variance: 1472.219

Conduct a hypothesis test about the mean value of the closing prices of the 25 different stocks you chose. The test you are conducting is trying to prove if the mean closing price of the stocks you chose is significantly different from $50. Conduct the test at the .05 level of significance. Be sure to state all assumptions behind the test conducted.Include sigma (ST DEV.), alpha (0.05), z value, p-value, and determine is p-value is <alpha. Do you reject the null or not?

Based on the results of your hypothesis test, you must then apply these results to appropriately analyze the situation presented. What do the conclusions of your test tell you practically? How might you use this information in the stock market or in making personal decisions?

Explanation / Answer

We are given that

mean: 34.841

St Dev: 38.36951

Variance: 1472.219 and N = 25

now we need to test that whether mean is 50 or not

so Hypothesis is

H0 : Mean = 50

H1 : Mean =| 50

Hence we calculate the t stat as

T = (mean1-mean2) / (sd/sqrt(n))

(50-34.84)/(38.36/sqrt(25)) = 1.94 , so our t stat is 1.94

now we must calculate t critical for df = n-1 = 25-1 = 24 and alpha =0.05

so reading the t table we get the critical value as 2.06

now as our t stat is not greater than t critical , hence we fail to reject the null hypothesis and conclude that mean is not significantly different from 50

The financial implications of it can be covered here as this is a statistical forum

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