Question
Please help me tutor. Thank you. Please calculate these question by USING EXCEL 1. Abrah college recently released a flyer in an attempt to recruit
Please help me tutor. Thank you. Please calculate these question by USING EXCEL
1. Abrah college recently released a flyer in an attempt to recruit accounting students for their business school. In the flyer, Abrah college claims that the mean first year graduate salary of their accounting students is $50,000, whereas their biggest competition for accounting students, Brown College, have a first year graduate salary of only $45,000.
After reading Abrah's recruiting flyer, Brown College executives believe that their own graduate salaries might have been initially incorrectly stated and as a result have conducted their own study. Based on the sample below, at 1% level of significance is there evidence that Brown's first year accounting graduates salaries are not as claimed?
Brown College First Year Accounting Graduate Salaries | |||||||||
$60,250 | $36,250 | $45,000 | $58,250 | $39,750 | $59,750 | $55,250 | $57,250 | $57,750 | $54,750 |
$56,500 | $33,500 | $55,500 | $59,000 | $51,000 | $58,750 | $40,000 | $47,000 | $54,500 | $41,750 |
$58,000 | $45,500 | $40,750 | $43,000 | $47,750 | $38,250 | $34,000 | $34,500 | $35,500 | $35,750 |
$38,250 | $36,250 | $46,000 | $53,500 | $37,500 | $43,250 | $46,250 | $58,000 | $52,250 | $53,250 |
$58,750 | $44,500 | $39,250 | $31,000 | $53,250 | $31,000 | $49,250 | $48,500 | $59,500 | $54,000 |
$44,000 | $48,750 | $33,250 | $60,250 | $32,500 | $60,000 | $43,250 | $36,750 | $53,250 | $55,750 |
$55,000 | $51,250 | $44,750 | $54,750 | $53,500 | $60,750 | $38,250 | $60,000 | $31,500 | $47,000 |
$54,000 | $49,250 | $46,250 | $58,000 | $34,500 | $36,250 | $44,250 | $31,000 | $42,750 | $59,750 |
$47,000 | $54,250 | $51,750 | $36,500 | $34,500 | $56,750 | $40,750 | $44,750 | $30,750 | $37,250 |
$36,000 | $57,250 | $54,000 | $36,000 | $47,750 | $50,500 | $47,500 | $44,000 | $51,750 | $40,500 |
$52,500 | $42,250 | $32,500 | $58,000 | $43,750 | $53,500 | $59,000 | $36,750 | $50,750 | $42,250 |
$31,250 | $54,750 | $33,000 | $30,750 | $30,000 | $44,000 | $35,250 | $58,000 | $61,000 | $39,500 |
$46,750 | $52,000 | $44,250 | $38,500 | $59,000 | $43,000 | $42,750 | $44,750 | $44,250 | $52,250 |
$34,250 | $57,250 | $52,000 | $48,000 | $30,000 | $50,000 | $55,500 | $59,000 | $56,250 | $49,500 |
$41,750 | $61,000 | $53,500 | $30,250 | $45,250 | $34,250 | $58,750 | $60,750 | $51,500 | $54,750 |
$56,500 | $39,250 | $49,250 | $49,750 | $59,000 | $52,750 | $42,750 | $40,750 | $49,500 | $31,500 |
$59,250 | $44,000 | $55,250 | $34,750 | $43,500 | $59,500 | $38,250 | $58,250 | $51,250 | $58,250 |
$42,750 | $56,500 | $46,750 | $60,500 | $51,500 | $59,000 | $53,500 | $46,750 | $56,000 | $37,750 |
$36,000 | $42,250 | $60,250 | $56,000 | $49,000 | $47,250 | $51,250 | $48,250 | $44,500 | $34,250 |
$35,750 | $42,500 | $53,750 | $34,250 | $32,250 | $47,250 | $55,000 | $32,750 | $57,250 | $33,750 |
$55,500 | $41,500 | $45,250 | $58,500 | $34,000 | $42,250 | $43,750 | $41,000 | $46,750 | $38,500 |
$36,500 | $46,750 | $49,250 | $42,000 | $55,000 | $34,000 | $60,250 | $57,750 | $47,250 | $57,250 |
BLANK #1: Is this a question involving mean or proportion and why ? ***ANSWER "MEAN" OR "PROPORTION"
BLANK #2: Which type of distribution should be used to calculate the probability for this question ? ***ANSWER "NORMAL", "T", OR "BINOMIAL"
BLANK #3: Which of the following options are the appropriate hypotheses for this question and why
A) H0: = $45,000 H1: > $45,000
B) H0: = $45,000 H1: < $45,000
C) H0: = $45,000 H1: $45,000
D) H0: p = $45,000 H1: p > $45,000
E) H0: p = $45,000 H1: p < $45,000
F) H0: p = $45,000 H1: p $45,000
BLANK #4: What is the p-value of this sample? ***ANSWER TO 4 DECIMALS
BLANK#5: Based on this sample, at 1% level of significance is there evidence that Brown's first year accounting graduates salaries are not as claimed and why? ***ANSWER "YES" OR "NO"
2. A company that produces cell phones claims its standard phone battery has a mean lifetime of 35 hours with a standard deviation of 2.56 hours. In order to investigate the company's claim, a consumer advocacy group takes a sample of 75 batteries and finds a mean lifetime of 34.28 hours. At a 1% level of significance, is there evidence for the consumer advocacy group to conclude that the mean lifetime of the batteries is lower than claimed?
BLANK #1: Is this a question involving mean or proportion? ***ANSWER "MEAN" OR "PROPORTION" and explain why
BLANK #2: Which type of distribution should be used to calculate the probability for this question and why? ***ANSWER "NORMAL", "T", OR "BINOMIAL"
BLANK #3: Which of the following options are the appropriate hypotheses for this question:
A) H0: = 34.28 hrs H1: > 34.28 hrs
B) H0: = 35 hrs H1: > 35 hrs
C) H0: = 34.28 hrs H1: < 34.28 hrs
D) H0: = 35 hrs H1: < 35 hrs
E) H0: = 34.28 hrs H1: 34.28 hrs
F) H0: = 35 hrs H1: 35 hrs
BLANK #4: What is the p-value of this sample? ***ANSWER TO 4 DECIMALS
BLANK#5: Based on this sample,at a 1% level of significance, is there evidence for the consumer advocacy group to conclude that the mean lifetime of the batteries is lower than claimed and explain?***ANSWER "YES" OR "NO"
3.The manufacturer of a certain brand of cigarettes states that the nicotine content in their cigarettes is 18.2mg with a standard deviation of 1.15mg. An independent testing agency examined a random sample cigarettes (sample below). At a 5% level of significance, is there evidence for the testing agency to conclude that the mean nicotine level to be higherthan the company states?
Cigarette Nicotine Level (mg) | ||||||||||||
18.41 | 18.58 | 17.19 | 18.22 | 18.94 | 17.11 | 18.71 | 18.35 | 18.21 | 17.74 | 19.29 | 18.11 | 17.72 |
17.07 | 19.14 | 17.32 | 19.6 | 18.05 | 18.32 | 17.07 | 19.41 | 18.03 | 19.29 | 17.13 | 18.51 | 18.57 |
18.18 | 18.66 | 18.27 | 19.01 | 17.12 | 18.85 | 17.85 | 17.34 | 19.46 | 17.66 | 19.23 | 19.38 | 17.4 |
19.12 | 17.03 | 18.92 | 17.89 | 17.26 | 17.8 | 18.87 | 18.21 | 18.81 | 17.36 | 18.72 | 17.09 | 18.82 |
19.09 | 19.35 | 18.74 | 19.26 | 17.27 | 18.93 | 18.43 | 18.93 | 19.07 | 18.7 | 17.99 | 18.84 | 18.05 |
19.44 | 17.53 | 17.01 | 19.5 | 18.57 | 18.35 | 17.53 | 19.21 | 17.54 | 18.61 | 17.79 | 18.55 | 19 |
18.55 | 18.29 | 19.45 | 18.49 | 18.78 | 19.31 | 18.76 | 19.12 | 18.96 | 18.07 | 17.35 | 18.73 | 18.51 |
19.57 | 18.77 | 19.08 | 17.08 | 18.62 | 17.07 | 18.53 | 17.55 | 19.43 | 17.24 | 19.19 | 17.79 | 17.87 |
17.17 | 19.03 | 17.94 | 18 | 17.34 | 19.32 | 18.94 | 17.22 | 18.75 | 18.43 | 18.14 | 17.21 | 17.83 |
19.34 | 17.74 | 17.81 | 17.9 | 17.67 | 18.51 | 19.53 | 17.59 | 19.12 | 18.83 | 17.88 | 18.76 | 17.44 |
17.98 | 19.46 | 19.5 | 18.15 | 17.82 | 17.33 | 18.4 | 17.79 | 18.47 | 19.52 | 17.45 | 18.77 | 18.73 |
19.23 | 18.94 | 18.23 | 17.95 | 17.44 | 19.3 | 17.74 | 19.27 | 17.52 | 19.3 | 18.46 | 18.01 | 19.18 |
17.07 | 18.27 | 17.91 | 19.15 | 18.61 | 18.99 | 19.11 | 19.39 | 18.55 | 17.52 | 17.13 | 19.19 | 19.21 |
18.4 | 17.37 | 18.07 | 18.47 | 19 | 17.5 | 17.06 | 17.13 | 17.18 | 18.97 | 17.37 | 17.18 | 17.4 |
17.55 | 17.15 | 17.21 | 18.04 | 18.83 | 17.81 | 19.12 | 19.3 | 18.17 | 18.91 | 17.76 | 19.45 | 18.48 |
17.65 | 19.4 | 17.18 | 18.05 | 17.75 | 19.26 | 17.9 | 18.43 | 19.6 | 17.26 | 19.16 | 17.53 | 17.91 |
17.4 | 17.76 | 17.4 | 17.67 | 18.96 | 17.67 | 17.12 | 17.82 | 17.4 | 19.46 | 18.22 | 18.35 | 19.21 |
18.35 | 17.34 | 18.96 | 17.85 | 17.51 | 18.17 | 17.91 | 19.23 | 18.91 | 19.32 | 18 | 17.87 | 17.16 |
18.51 | 17.55 | 19.11 | 17.86 | 18.46 | 19.15 | 18.91 | 18.7 | 18.94 | 19.14 | 18.09 | 17.24 | 18.8 |
18.76 | 18.28 | 17.23 | 17.04 | 17.21 | 18.39 | 18.6 | 18.48 | 17.04 | 18.9 | 18.57 | 18.48 | 17.79 |
19.57 | 18.91 | 18.83 | 18.74 | 18.47 | 17.26 | 19.6 | 18.91 | 18.02 | 19.18 | 18.04 | 18.66 | 19.41 |
BLANK #1: Is this a question involving mean or proportion and why? ***ANSWER "MEAN" OR "PROPORTION"
BLANK #2: Which type of distribution should be used to calculate the probability for this question and why? ***ANSWER "NORMAL", "T", OR "BINOMIAL"
BLANK #3: Which of the following options are the appropriate hypotheses for this question:
A) H0: = 18.2mg H1: > 18.2mg
B) H0: = 18.2mg H1: < 18.2mg
C) H0: = 18.2mg H1: 18.2mg
D) H0: p = 18.2mg H1: p > 18.2mg
E) H0: p = 18.2mg H1: p < 18.2mg
F) H0: p = 18.2mg H1: p 18.2mg
BLANK #4: What is the p-value of this sample? ANSWER TO 4 DECIMALS
BLANK#5: Based on this sample,at a 5% level of significance, is there evidence for the testing agency to conclude that the mean nicotine level to be higherthan the company states? ***ANSWER "YES" OR "NO" and explain
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