test #3 cheat sheet

# test #3 cheat sheet - N= Number sampled Sx= = median X= X(x...

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N= Number sampled Sx= µ= median X= X (x bar)= sample statistic α = level of significance Zc= level of confidence Alpha Tail Z 0.10 Left Right Two - 1.28 +1.28 ± 1.645 0.05 Left Right Two - 1.645 +1.645 ± 1.96 0.01 Left Right two -2.33 +2.33 ±2.575 Left tail FAIL TO H0: μ≥k Reject Ho REJECT Ho Ha : μ<k Right Tail FAIL TO Reject Ho H0: μ≤k REJECT Ho Ha : μ>k ALWAYS 1-Z VALUE to get FAIL TO P-VALUE Reject Ho REJECT Ho Reject Ho Two Tailed H0: μ=k If P α , then reject H 0 . Ha : μ≠k If P > , then fail to reject H 0 . Z-TEST ON CALC Null hypothesis - H 0 : A statistical hypothesis that contains a statement of equality such as , =, or . Alternative hypothesis - H a : A statement of inequality such as >, , or < Ex Find the critical value (t o ) for a left tail test α=0.025, n=19 d.f= (n-1) = 19-1=18 (use t-table with d.f. =18 and 0.025) t o = -2.101 ( left tail means - ) Ex Find the critical value (t o ) for a two tail test α=0.01, n=27 d.f= (n-1) = 27-1=26 (use t-table with d.f. =26 and 0.01) t o = ±2.779 ( two tail means ± ) Ex Find the critical value (t o ) for a right tail test α=0.10, n=20 d.f= (n-1) = 20-1=19 (use t-table with d.f. =19 and 0.10) t o =1.328 ( right tail means + ) Ex state whether the standardized test statistic t indicates that you should reject the null Hypothesis. T o = -2.086 Ex The Pew Research Center claims that more than 30% of US adults regularly watch the weather channel. You decide to test this claim and ask a random sample of 75 adults whether they regularly watch the weather channel. Of the 75 adults, 27 responded yes. At α=0.01, is there enough evidence to support the claim? N=75 X= 27 P=0.30 NP= 75(.3) =22.5 YES Ho P≤ 0.30 Q=0.70 NQ= 75(.7) =52.5 Ha P> 0.30 (claim) α=0.01 Critical value z=1.13 α=0.01 Zo= 2.33 fail to reject Ho Zo=2.33 Rejection region Z >2.33 (in alpha chart) P = 27/25 = 0.36 (0.36-0.30) / √(0.30)(0.7)/75 = Z= 1.13 Fail to reject Ho. Not enough evidence to support the claim. Ex A company specializing in parachute assembly state that it’s main parachute failure rate is not more than 1%. Perform a hypothesis test to determine whether the company’s claim is false. P= failure rate Ho: P≤ 0.01 (claim) Ha: P>0.01 Type 1 error: reject Ho P≤0.01 when actually P≤ 0.01 Type 2 error: fail to reject Ho P≤0.01 When actually P≤0.01 Ex Find the p-value for a left tailed hypothesis test with a test statistic of z= -1.62. Decide whether to reject Ho if the level of significance is α=0.05 Ho µ ≥K Ha µ <K z= -1.62 P= 0.0526 P Value = 0.0526 (z on chart) P Value= 0.0526 > 0.05 =α Fail to reject H o Ex find the p-value for a two tailed hypothesis test with a test statistic of z=2.31. Decided whether to reject Ho if the lever of significance is

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test #3 cheat sheet - N= Number sampled Sx= = median X= X(x...

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