4 inference questions
A random sample of 41 observations was selected from a normally distributed population. The
sample mean was = 55, and the sample variance was s2 = 33.0. Does the sample show sufficient
reason to conclude that the population standard deviation is not equal to 8 at the 0.05 level of
significance? Use the p-value method. · State the null and alternate hypotheses · Determine which
test statistic applies, and calculate it · Determine the corresponding probability, and compare to ·
State your decision: Should the null hypothesis be rejected?
n = 41
xbar = 55
s^2 = 33.0
H0: σ =8
H1: σ≠8
α = 0.05
The chi-square test for variance is to be applied.
chi-square = (n-1)s^2/σ^2 = 40 * 33.0/8^2 = 40*33.0/64 = 20.625
Df = (n-1) = 41-1=40
The p-value is 0.009594
Since the p-value is less than the significance level 0.05, we reject the null hypothesis.
An insurance company states that 75% of its claims are settled within 5 weeks. A consumer group
selected a random sample of 45 of the company’s claims and found 35 of the claims were settled
within 5 weeks. Is there enough evidence to support the consumer group’s claim that fewer than
75% of the claims were settled within 5 weeks? Test using the traditional approach with a = 0.05. ·
State the null and alternate hypotheses · Calculate the sample proportion · Determine which test
statistic applies, and calculate it · Determine the critical value(s). · State your decision: Should the
null hypothesis be rejected? .
H0: P = 0.75
H1: P < 0.75
Sample proportion P^ = 35/45 =0.778
The z test for proportions is to be used.
z = (P^ - P)/√(P*(1-P)/n) = (0.778 – 0.75)/√(0.75*0.25/45) = 0.430
The critical value of z is -1.65
The critical region is z < -1.65
The computed value of z = 0.430 does not fall in the critical region.
The null hypothesis cannot be rejected.
There is not enough evidence to support the consumer group’s claim that fewer than 75% of the
claims were settled within 5 weeks.
A teacher wishes to compare two different groups of students with respect to their mean time to
complete a standardized test. The time required is determined for each group. The data summary is
given below. Test the claim at = 0.10, that there is no difference in variance. Give the critical region,
test statistic value, and conclusion for the F test. n1 = 60 s1 = 30 n2 = 120 s2 = 40 = 0.10 · State the
null and alternate hypotheses · Determine which test statistic applies, and calculate it · Determine
the critical region · State your decision: Should the null hypothesis be rejected?
H0: σ1^2 = σ2^2
H1: σ1^2 ≠ σ2^2
The F test for equality of variances is to be applied.
F = s1^2/s2^2 = 30^2/40^2 = 900/1600 = 0.5625
Degrees of freedom of F = (n1-1, n2-1) = (59,119)
The critical values of F are 0.680 and 1.432
The critical region is
F < 0.680 or F > 1.432
The computed value of F is in the critical region. So the null hypothesis is rejected.
The data does not support the claim that the variances are the same.
The pulse rates below were recorded over a 30-second time period, both before and after a physical
fitness regimen. The data is shown below for 8 randomly selected participants. Is there sufficient
evidence to conclude that a significant amount of improvement took place? Assume pulse rates are
normally distributed. Test using = 0.10. · State the null and alternate hypotheses · Calculate the
mean and standard deviation · Determine which test statistic applies, and calculate it · Determine
the critical value(s). · State your decision: Should the null hypothesis be rejected? Before 31 31 46 31
52 45 59 44 After 32 31 51 33 52 46 63 45
H0: µ1 = µ2 (or µ2 -µ1 = 0)
H1: µ1 < µ2 (or µ2 - µ1 > 0)
Differences of the measurements are
1
0
5
2
0
1
4
n=8
Sum of the differences = Σd = 14
Mean of the differences = dbar = 14/8 = 1.75
Sum of squares of the differences = 48
Variance of the differences = (48 – 14^2/8)/(8-1) = 3.357
Standard deviation of the differences =sd = sqrt(3.357) =1.832
The paired t test for equality of means is to be applied.
t = √n (dbar -0)/sd = √8 (1.75 – 0)/1.832 = 2.701
Df = 8-1 = 7
The critical value of t is 1.895
The critical region is t > 1.895
The computed value of t falls in the critical region.
So the null hypothesis is rejected.
There is sufficient evidence to conclude that a significant amount of improvement took place.
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