Question

Solve in excel please. Show formulas

A manufacturer of raisin bran cereal claims that each box of cereal has more than 200 grams of

raisins. The firm selects a random sample of 64 boxes and records the amount of raisin (in grams) in

each box.

a. Identify the null and the alternate hypotheses for this study.

b. Is there statistical support for the manufacturer’s claim at a significance level of 5%? What

about at 1%? Test your hypothesis using both, the critical value approach and the p-value

approach. Clearly state your conclusions.

c. Under what situation would a Type-I error occur? What would be the consequences of a

Type-I error?

d. Under what situation would a Type-II error occur? What would be the consequences of a

Type-II error?

Box Amount

1 140

2 310

3 .276

4 174

5 136

6 272

7 376

8 324

9 252

10 84

11 176

12 250

13 177

14 89

15 254

16 185

17 186

18 94

19 94

20 221

21 211

22 308

23 169

24 217

25 363

26 123

27 259

28 110

29 102

30 134

31 295

32 171

33 94

34 331

35 218

36 158

37 213

38 244

39 166

40 216

41 156

42 360

43 198

44 217

45 246

46 256

47 258

48 374

49 338

50 276

51 212

52 216

53 168

54 376

55 245

56 252

57 373

58 270

59 245

60 108

61 190

62 208

63 231

64 206

Answer 1

Given that,
population mean(u)=200
sample mean, x =221.109
standard deviation, s =78.828
number (n)=64
null, Ho: μ=200
alternate, H1: μ>200
level of significance, α = 0.05
from standard normal table,right tailed t α/2 =1.669
since our test is right-tailed
reject Ho, if to > 1.669
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =221.109-200/(78.828/sqrt(64))
to =2.1423
| to | =2.1423
critical value
the value of |t α| with n-1 = 63 d.f is 1.669
we got |to| =2.1423 & | t α | =1.669
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :right tail - Ha : ( p > 2.1423 ) = 0.01802
hence value of p0.05 > 0.01802,here we reject Ho
ANSWERS
---------------
a.
null, Ho: μ=200
alternate, H1: μ>200
b.
i.
level of significance =0.05
test statistic: 2.1423
critical value: 1.669
decision: reject Ho
p-value: 0.01802
we have enough evidence to support the claim that manufacturer of raisin bran cereal claims that each box of cereal has more than 200 grams of
raisins.
ii.
level of significance =0.01
Given that,
population mean(u)=200
sample mean, x =221.109
standard deviation, s =78.828
number (n)=64
null, Ho: μ=200
alternate, H1: μ>200
level of significance, α = 0.01
from standard normal table,right tailed t α/2 =2.387
since our test is right-tailed
reject Ho, if to > 2.387
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =221.109-200/(78.828/sqrt(64))
to =2.1423
| to | =2.1423
critical value
the value of |t α| with n-1 = 63 d.f is 2.387
we got |to| =2.1423 & | t α | =2.387
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value :right tail - Ha : ( p > 2.1423 ) = 0.01802
hence value of p0.01 < 0.01802,here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=200
alternate, H1: μ>200
test statistic: 2.1423
critical value: 2.387
decision: do not reject Ho
p-value: 0.01802
we do have enough evidence to support the claim that manufacturer of raisin bran cereal claims that each box of cereal has more than 200 grams of
raisins.
c.
in part (b),
i.
we reject the null hypothesis so that type 1 error is possible for this situation.
Given that,
Standard deviation, σ =78.828
Sample Mean, X =221.109
Null, H0: μ=200
Alternate, H1: μ>200
Level of significance, α = 0.05
From Standard normal table, Z α/2 =1.64
Since our test is right-tailed
Reject Ho, if Zo < -1.64 OR if Zo > 1.64
Reject Ho if (x-200)/78.828/√(n) < -1.64 OR if (x-200)/78.828/√(n) > 1.64
Reject Ho if x < 200-129.278/√(n) OR if x > 200-129.278/√(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 64 then the critical region
becomes,
Reject Ho if x < 200-129.278/√(64) OR if x > 200+129.278/√(64)
Reject Ho if x < 183.84 OR if x > 216.16
Suppose the true mean is 221.109
Probability of Type I error,
P(Type I error) = P(Reject Ho | Ho is true )
= P(183.84 < x OR x >216.16 | μ1 = 221.109)
= P(183.84-221.109/78.828/√(64) < x - μ / σ/√n OR x - μ / σ/√n >216.16-221.109/78.828/√(64)
= P(-3.782 < Z OR Z >-0.502 )
= P( Z <-3.782) + P( Z > -0.502)
= 0.0001 + 0.6922 [ Using Z Table ]
= 0.692
d.
in part (b),
ii.
we fails to reject the null hypothesis so that type 2 error is possible for this condition.
Given that,
Standard deviation, σ =78.828
Sample Mean, X =221.109
Null, H0: μ=200
Alternate, H1: μ>200
Level of significance, α = 0.01
From Standard normal table, Z α/2 =2.3263
Since our test is right-tailed
Reject Ho, if Zo < -2.3263 OR if Zo > 2.3263
Reject Ho if (x-200)/78.828/√(n) < -2.3263 OR if (x-200)/78.828/√(n) > 2.3263
Reject Ho if x < 200-183.3776/√(n) OR if x > 200-183.3776/√(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 64 then the critical region
becomes,
Reject Ho if x < 200-183.3776/√(64) OR if x > 200+183.3776/√(64)
Reject Ho if x < 177.0778 OR if x > 222.9222
Implies, don't reject Ho if 177.0778≤ x ≤ 222.9222
Suppose the true mean is 221.109
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(177.0778 ≤ x ≤ 222.9222 | μ1 = 221.109)
= P(177.0778-221.109/78.828/√(64) ≤ x - μ / σ/√n ≤ 222.9222-221.109/78.828/√(64)
= P(-4.4686 ≤ Z ≤0.184 )
= P( Z ≤0.184) - P( Z ≤-4.4686)
= 0.573 - 0 [ Using Z Table ]
= 0.573
For n =64 the probability of Type II error is 0.573