At a certain university, 22% of the students fail general chemistry on their first attempt. Professor Brown teacher at this university and believes that the rate of first-time failure in his general chemistry classes is 45%. He samples 86 students from last semester who were first-time enrollees in general chemistry and finds that 19 of them failed his course.
1) State the appropriate null and alternate hypotheses.
2) Compute the P-value.
3) Using a = 0.05, can Professor Brown conclude that the percentage of failures differs from 45%?

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opudodennis opudodennis · Answer 1 · 2020-04-12T05:39:55+02:00

Answer:

[tex]a. \ H_o:p=0.45, \ \ \ \ H_a:p\neq 0.45\\\\b.\ \hat p=0.2209\\\\c. \ Yes\ (-4.2706<-1.96)[/tex]

Step-by-step explanation:

a. The professor's claim is that 45% first-timers fail his test. The null hypothesis is therefore stated as:

[tex]H_o:p=0.45[/tex]

-The alternative hypothesis is that more or less people fail the test as opposed to the professor's exact claim, hence:

[tex]H_a:p\neq 0.45[/tex]

b. To compute the P-value we use the z-value for a 95% confidence level:

[tex]z_{\alpha/2}=z_{0.025}=1.96[/tex]

#The proportion of failures in the sample of 86 is 19:

[tex]\hat p=\frac{19}{86}\\\\=0.2209[/tex]

The z-value is calculated as:

[tex]z=\frac{\hat p-p_o}{\sqrt{\frac{p_o(1-p_o)}{n}}}[/tex]

[tex]=\frac{0.2209-0.45}{\sqrt{\frac{0.45(1-0.45)}{86}}}\\\\\\=-4.2706[/tex]

-4.2706 is less than the stated confidence level for the given 45% proportion and greatly differs from it.

- Reject the null hypothesis as there is enough evidence to reject the claim.

-Hence,Yes, Professor Brown can conclude that percentage of failures differs from 45%.