What is the elements of the hypothesis testing?

Question

Statistics students believe that the mean score on a first statistics test is 65. The instructor thinks that the mean score is higher. She samples 10 statistics students and obtains the scores:

Grades

88

74.4

64.3

68.4

61.9

74.4

96

61.9

85.5

88

Use a significance level of 5%.

1. State the alternative hypothesis:
2. State the mean of the sample
3. State the standard error of the sample means
4. State the test statistic: t= 
5. State the p-value
6. Decision=

Answer 1

Hi Kaitlyn,

Null hypothesis (H₀) means exactly that—null, as in nothing. In this case, no difference in mean. Alternative hypothesis (H_A) is the opposite of that—essentially something. In this case:

1. H₀: Mean score on first statistics test does not differ significantly from 65.

H_A: Mean score on first statistics test is significantly greater than 65.

1. 
   
2. To get sample mean x-bar, add up all test scores and divide by sample size, 10.

X-bar=(88+74.4+64.3+68.4+61.9+74.4+96+61.9+85.5+88)/10

X-bar=76.28

3.

Standard error is computed as:

SE=s/sqrt(n)

s=sample standard deviation

n=sample size

Use a calculator, software, or online calculator to get sample standard deviation.

s=12.36

n=10

SE=12.36/sqrt(10)

SE=3.91

(4) Sorry about confusing numbering; Wyzant text editor is not flexible with this—it keeps resetting to 1. Anyhow, to compute t-test statistic:

t=(x-bar- mu₀)/SE

Breaking this down:

x-bar=sample mean

mu₀=hypothesized mean

SE=standard error

Thus:

x-bar=76.28

mu₀=65

SE=3.91

t=(76.28-65)/3.91

t=2.88

(5) To approximate p-value, we need degrees of freedom, which is computed as:

df=n-1

df=10-1

df=9

We also need our confidence level, 95%. Keep in mind this is a one-sided test. Proceed to t-table. Look for row with 9 degrees of freedom and look in interior for t-statistic computed above. It falls between 2.821 and 3.25. Look at top row to see what corresponding p-values are. Look for one-sided test.

This gives:

0.005<p<0.01

This is all we can get for p-value without statistical software.

(6)

From above, p is clearly less than 0.05, which is the cutoff for significance at the 95% confidence level. This means we can reject H₀ and conclude that true mean score on first statistics test significantly exceeds 65.

I hope this helps.