Please answer in detail Ius Data Review View Tell me what you want to do P56 30
ID: 3332589 • Letter: P
Question
Please answer in detail
Ius Data Review View Tell me what you want to do P56 30 31 According to Thomson Financial, through January 25, 2006, the majority 4th quarter of 2012, 191 of a random sample of 300 companies reported profits that hacd f companies reporting profits had beaten estimates. Now suppose that for the beaten estimates. Use this latest information to answer the following questions. (See exercise 38 on page 369 of your textbook for a similar problem.) 35 Does the data show the proportion of companies in the 4th quarter of 2012 that beat estimates is greater than 57% at a-0.1? 37 38 Question 2A Step 1 Ho: 0.57 40 41 42 0.57 Step 2 45 46 47 48 Fill in row 48 if the hypothesis is one tailed. Reject Ho if Fill in row 52 if the hypothesis is two tailed. The smaller number must be typed first. Reject Ho if Step 3: 51 2.45 or t stat t stat 53 Step 4 Step 5 Step 6 Step 7 56 57 59 Fill in row 62 it the Ztable is appropriate. 61 p value 63 Fill in rowe 66 if the t toble is appropriste The lower bouridary must go on the loft and the upper boundary on the righ P value 66 1 67 HWNotes HWExcelDirections HW Survey DishonestyExplanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: P < 0.57
Alternative hypothesis: P > 0.57
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected only if the sample proportion is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.10. The test method, shown in the next section, is a one-sample z-test.
Analyze sample data. Using sample data, we calculate the standard deviation () and compute the z-score test statistic (z).
= sqrt[ P * ( 1 - P ) / n ]
= 0.02858
z = (p - P) /
z = 2.33
zcritical = 1.645
If zvalue greater than 1.645, reject the null hypothesis.
where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.
Since we have a one-tailed test, the P-value is the probability that the z-score is more than 2.33. We use the Normal Distribution Calculator to find P(z > 2.33) = 0.0099
Thus, the P-value = 0.0099
Interpret results. Since the P-value (0.0099) is less than the significance level (0.10), we cannot accept the null hypothesis.