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5) x̅ = 71.2, s = 2.4, n = 25

α = 0.05

Null and Alternative hypothesis:

Ho : µ = 69.5 (Claim)

H1 : µ ≠ 69.5

Test statistic:

t = (x̅- µ)/(s/√n) = (71.2 - 69.5)/(2.4/√25) = 3.5417

df = n-1 = 24

Critical value :

Two tailed critical value, t-crit = T.INV.2T(0.05, 24) = 2.064

Or using p-value :

Two tailed p-value = T.DIST.2T(ABS(3.5417), 24) = 0.0017

Decision:

p-value < α, Reject the null hypothesis.

Conclusion:

There is not enough evidence to support the claim that the mean height is 69.5 at 0.05 significance level.

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6). n = 400, x = 244

p̄ = x/n = 0.61

α = 0.05

Null and Alternative hypothesis:

Ho : p ≥ 0.70 (Claim)

H1 : p < 0.70

Test statistic:

z =(p̄ -p)/(√(p*(1-p)/n)) = -3.9279

Critical value :

Left tailed critical value, z crit = NORM.S.INV(0.05) = -1.645

Or using p-value :

p-value = NORM.S.DIST(-3.9279, 1) = 0.0000

Decision:

p-value < α, Reject the null hypothesis.

There is not enough evidence to support the claim that the proportion is at least 70% at 0.05 significance level.

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7).

Category	Observed Frequency (O)	Expected Frequency (E)	(O-E)²/E
A	25	210 * 0.14 = 29.4	(25 - 29.4)²/29.4 = 0.6585
B	48	210 * 0.2 = 42	(48 - 42)²/42 = 0.8571
C	63	210 * 0.3 = 63	(63 - 63)²/63 = 0
D	43	210 * 0.2 = 42	(43 - 42)²/42 = 0.0238
F	31	210 * 0.16 = 33.6	(31 - 33.6)²/33.6 = 0.2012
Total	210	210	1.7406

Null and Alternative hypothesis:

Ho: p1 = 0.14, p2 = 0.20, p3 = 0.30, p4 = 0.20, p5 = 0.16 (Claim)

H1: Proportions are different.

Test statistic:

χ² = ∑ ((fo-fe)²/fe) = 1.7406

df = n-1 = 4

p-value:

p-value = CHISQ.DIST.RT(1.7406, 4) = 0.7833

Decision:

p-value > α, Do not reject the null hypothesis

There is enough evidence to support the teacher claim at 0.05 significance level.