Test Statistic / Z-Test vs. T-Test Question?
Since we are testing for proportion, a z-test is OK. Also, we have a large sample. a) H0: Median income in 2003 is $52,000 ( p=0.5) HA: Median income is over $52,000 (p > 0.5) p = proportion of people with income over $52,000 in 2003 b) Sample proportion phat = 0.56 Variance of proportion = p*(1-p)/n = 0.5(0.5)/750 =0.0003333 S.D. of p is sqrt[0.000333] = 0.0183 Z = ( 0.56 - 0.5 ) / 0.0183 = 3.2863 c) P( z > 3.29) = 0.00001 The critical z assuming a 5 % level of significance is 1.645. 3.29 > 1.645; hence, reject H0 and accept HA. The proportion of people with median income $52,000 is greater than 50 %. Note: We need to know in 1999, what percentage of people had income over $52,000 to precisely do a statistical test. (or at least a sample estimate taken in 1999).
For a test of Ho: p = 0.5, the z test statistic equals -1.52. Find the p-value for Ha: p < 0.5...?
The p-value of the test is the area under the normal curve that is in agreement with the alternate hypothesis. H1: p > p0; p-value is the area to the right of z H1: p < p0; p-value is the area to the left of z H1: p ≠ p0; p-value is the area in the tails greater than |z| here we have: p-value = P(Z < -1.52) = 0.06425549
College Statistics question - hypothesis testing?
ANSWER: Conclusion: H1 is true Why???? SINGLE SAMPLE TEST, ONE-TAILED, 6 - Step Procedure for t Distributions, "one-tailed test" Step 1: Determine the hypothesis to be tested. Lower-Tail H0: μ ≥ μ0 H1: μ < μ0 or Upper-Tail H0: μ ≤ μ0 H1: μ > μ0 hypothesis test (lower or upper) = upper Step 2: Determine a planning value for α [level of significance] = 0.1 Step 3: From the sample data determine x-bar, s and n; then compute Standardized Test Statistic: t = (x-bar - μ0)/(s/SQRT(n)) x-bar: Estimate of the Population Mean (statistical mean of the sample) = 3.62 n: number of individuals in the sample = 40 s: sample standard deviation = 0.17 μ0: Population Mean = 3.54 significant digits = 3 Standardized Test Statistic t = ( 3.62 - 3.54 )/( 0.17 / SQRT( 40 )) = 2.976 Step 4: Use Students t distribution, 'lookup' the area to the left of t (if lower-tail test) or to the right of t (if upper-tail test) using Students t distribution Table or Excel TDIST(x, n-1 degrees_freedom, 1 tail) =TDIST( 2.976 , 39 , 1 ) Step 5: Area in Step 4 is equal to P value [based on n -1 = 39 df (degrees of freedom)] = 0.002 Table look-up value shows area under the 39 df curve to the right of t = 2.976 is (approx) probability = 0.002 Step 6: For P ≥ α, fail to reject H0; and for P < α, reject H0 with 90% confidence. Conclusion: H1 is true Note: level of significance [α] is the maximum level of risk an experimenter is willing to take in making a "reject H0" or "conclude H1" conclusion (i.e. it is the maximum risk in making a Type I error).
Z-test and z critical value statistics question. how do you know whether to reject a null hypothesis?
There are TWO methods for taking decision. 1) Comparing the test statistic with the critical value (of z , t , chi-square...) REJECT the Ho if the test statistic > the critical value FAIL TO REJECT or ACCEPT the Ho if the test statistic < the critical value 2) Comparing the p value with alpha value REJECT the Ho if the p value < alpha value (0.01 , 0.05, 0.10 ...) FAIL TO REJECT or ACCEPT the Ho if the p value > alpha value Therefore, in your case the decision is ACCEPT the Ho.
For a test of Ho: p = 0.5, the z test statistic equals 1.52. Find the p-value for Ha: p < 0.5.?
I think what you are asking for is the probability of achieving a z-score of 1.52 or higher. I ran across a very similaqr question that may help.