Question 1
The null and alternate hypothesis will be;
Ho : µ ≥ 45
H1: µ < 45
Z-calculated = (x- µ)/ δM
Where;
x is the obtained mean score which is 36.4
µ is the mean that was hypothesized which is 45
δM is the standard error that is calculated by δ/√n = 4/√100 = 0.4
Hence;
Z-calculated = (36.4- 45)/0.4 = -21.5
It is a one tailed test since we are only interested as to whether the mean age has decreased. For a one tailed test, at α = 0.05; the z critical values are +/- 1.650.
The z obtained falls within the rejection region. We should reject the null hypothesis and accept the alternate hypothesis. Therefore, we can conclude that the population’s mean age of audience for American Idol has declined.
Question 2
The statistical test to be used is the t- test because n < 30
The null and alternate hypothesis will be;
Ho : µ ≥ 21.5
H1: µ < 21.5
t-calculated = (x- µ)/ δM
Where;
x is the obtained mean score which is 21
µ is the mean that was hypothesized which is 21.5
δM is the standard error that is calculated by δ/√n = 1.6/√20 = 0.3578
Hence;
t-calculated = (21- 21.5)/0.3578 = 1.397
The table value for t with 19 degrees of freedom (20-1) at 10% significance level is 1.328.
The t calculated falls within the rejection region. We should reject the null hypothesis and accept the alternate hypothesis. Therefore, we can conclude that the population’s mean uterine weight is less than 21.5 mg.
Works Cited
Healey, Joseph F. Statistics: A Tool for Social Research. revised. London: Cengage Learning, 2011.
