Answer:
Here we have given two catogaries as degree holder and non degree holder.
So here we have to test the hypothesis that,
H0 : p1 = p2 Vs H1 : p1 not= p2
where p1 is population proportion of degree holder.
p2 is population proportion of non degree holder.
Assume alpha = level of significance = 5% = 0.05
The test is two tailed.
Here test statistic follows standard normal distribution.
The test statistic is,
Z = (p1^ - p2^) / SE
where SE = sqrt[(p^*q^)/n1 + (p^*q^)/n2]
p1^ = x1/n1
p2^ = x2/n2
p^ = (x1+x2) / (n1+n2)
This we can done in TI_83 calculator.
steps :
STAT --> TESTS --> 6:2-PropZTest --> ENTER --> Input all the values --> select alternative "not= P2" --> ENTER --> Calculate --> ENTER
Test statistic Z = 1.60
P-value = 0.1090
P-value > alpha
Fail to reject H0 or accept H0 at 5% level of significance.
Conclusion : There is not sufficient evidence to say that the percent of correct answers is significantly different between degree holders and non-degree holders.
Answer:
3934
Step-by-step explanation:
Round 15,479,652 and find square root.
Answer:
B: 32 cm
Step-by-step explanation:
Add together only the outside lengths
6+6+4+4+3+3+3+3=32 cm
