NULL HYPOTHESIS H0: Medium= $1700
ALTERNATIVE HYPOTHESIS HA: Medium $1700
m > md, then we should expect more than half of the xi − m0 quantities obtained to be positive and fewer than half to be negative.
Number of negative observation x= 6
n= 15
p= 0.5
Therefore, we need to calculate how likely it would be to observe as few as 6 negative signs if the null hypothesis were true. Doing so, we get a P-value of 0.3036.
Since in the question alpha is not given. If we assume alpha=0.05
then P value is not significant. Therefore we do not reject null hypothesis at 0.05 level of significance.
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