Solution :
Given that ,
n = 203
x = 55
The null and alternative hypothesis is
H0 : p = 0.20
Ha : p > 0.20
This is the right tailed test .
= x / n = 55 / 203 = 0.2709
P0 = 20% = 0.20
1 - P0 = 1 - 0.20 = 0.8
Test statistic = z
= - P0 / [P0 * (1 - P0 ) / n]
= 0.2709 - 0.20 / [0.20 (1 - 0.20) / 203 ]
= 2.527
The test statistic for the hypothesis = 2.53
P-value = 0.0057
The P-value for the hypothesis = 0.006
= 0.05
0.006 < 0.05
P-value <
Reject the null hypothesis .
Conclusion : - B. Reject Ho.There is sufficient evidence to warrant support of the claim that more than 20% of users devers nausea.
Given that ,
The null and alternative hypothesis is
H0 : p = 0.402
Ha : p 0.402
This is the two tailed test .
Test statistic = z = 2.69
P-value = 0.0072
The P-value = 0.007
= 0.05
0.007 < 0.05
P-value <
Reject the null hypothesis .
Conclusion : - Reject Ho.There is sufficient evidence to support the claim that p 0.402
Suppose 203 subjects are treated with a drug that is used to treat pain and 55...
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