Answer:
Step-by-step explanation:
<u><em>the mean in period</em></u> 1 :
(2.3+2.1+2.2+2.2+2.2+2.1+2.4+2.5+2.2+2.0+1.9+1.9+2.1+2.2+2.3)÷15=21.733...
<u><em>the mean in period</em></u> 2 :
(2.3+2.1+3.3+1.5+3.6+1.6+3.0+1.1+4.7+2.1+2.4+1.9+2.8+0.5+2.3)÷15=23.466...
Since 23.466 > 21.733 then “The mean in period 2 is higher than the mean in period 1”.
Answer:
Step-by-step explanation:
The summary of the given data includes;
sample size for the first school = 42
sample size for the second school = 34
so 16 out of 42 i.e = 16 and 18 out of 34 i.e
= 18 have ear infection.
the proportion of students with ear infection Is as follows:
= 0.38095
= 0.5294
Since this is a two tailed test , the null and the alternative hypothesis can be computed as :
level of significance ∝ = 0.05,
Using the table of standard normal distribution, the value of z that corresponds to the two-tailed probability 0.05 is 1.96. Thus, we will reject the null hypothesis if the value of the test statistics is less than -1.96 or more than 1.96.
The test statistics for the difference in proportion can be achieved by using a pooled sample proportion.
The pooled standard error can be computed by using the formula:
The test statistics is ;
z = - 1.368
Decision Rule: Since the test statistics is greater than the rejection region - 1.96 , we fail to reject the null hypothesis.
Conclusion: There is insufficient evidence to support the claim that a difference exists between the proportions of students who have ear infections at the two schools