The test statistic in a two dash tailed test is z= -2.84. Determine the P-value and decide whether, at the 1% significance level, the data provide sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis.
*Please show me a way to do it on the calculator (ti-84)*
P value is 0.0045
P value is 0.0045 < 0.01
Conclusion: at the 1% significance level, the data provide sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis.
Steps by using TI-84 for p value.
Find the absolute value of z_calc>>2nd DISTR>> Scroll
down to normalcdf( >> ENTER
>>Now enter: |z_calc|, 1000, 0,1)
>> ENTER
>> Output is ½ of the P-value so
>> Multiply result by 2
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