Can you please help me solve this using a TI-83 Plus Calculator? I am having tro

Question

Can you please help me solve this using a TI-83 Plus Calculator? I am having trouble finding the Critical Value and P-Value.

Assume that a simple random sample has been selected from a normally distributed population and test the given claim. Use either the traditional method or P-value method as indicated. Identify the null and alternative hypotheses, test statistic, critical value(s) or P-value (or range of P-values) as appropriate, and state the final conclusion that addresses the original claim. 40) In tests of a computer component, it is found that the mean time between failures is 520 40) hours. A modification is made which is supposed to increase the time between failures Tests on a random sample of 10 modified components resulted in the following times (in hours) between failures 518 548 561 523 536 499 538 557 528 563 At the 0.05 significance level, test the claim that for the modified components, the mean time between failures is greater than 520 hours. Use the P-value method of testing hypotheses. Ho: Test statistic: We There P-value- the null hypothesis sufficient evidence to support the claim

Explanation / Answer

Here first i will tell you how to get critical value and p value from TI-83 Plus Calculator. I will solve the question later.

For critical t value.

S1 : Press the “STAT” key, then press the left arrow key to arrive at the Tests menu.

S2 : Press “8” for TInterval.

S3 :Here put sample mean as 0 and standard deviation as standard error. Here this step tell use about the interval for the given confidence level.

(a) So, if two tailed test, directly put C - level : 95 or whatever directly.

(b) IF one tailed test, then double the C - level i.e. 90% for 0.05 confidence level, 98% for 0.01 confidence level.

The value you will get is the critical t value.

Now for p value

(1) calculate t_calc (t - statistic)

(2) 2nd DSTR

(3) Scroll Down to tcdf (

(4) Enter

(5) Now enter : t _ Calc, 1000 , dF

(6) Enter

(7) Output is p -value

Now i will solve the question

H0 : <= 520

Ha : > 520

sample mean x = 537.1 ours

sample standard deviation s = 20.7013 hours

standard error of sample mean se0 = s/sqrt(n) = 20.7013 / sqrt(10) = 6.5463 hours

test statistic

t = (537.1 - 520)/ 6.5463 = 2.61

so here p - value = TDIST (t > 2.61; 9 ;one tailed) = 0.0141

so here p - value < 0.05 so we can reject the null hypothesis so we can conclude that there is significant higher than $ 520.

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