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Hyspothesis Testing

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Large Sample Mean Test

One data

Soal : Seorang peneliti melaporkan bahwa rata-rata gaji asisten profesor lebih dari $42.000. Sampel 30 asisten profesor memiliki rata-rata gaji $43.260. Pada α = 0,05, uji klaim bahwa asisten profesor mendapatkan lebih dari $42.000 setahun. Standar deviasi populasi adalah $5230.
Solusi :
1

State the hypotheses and identify the claim

H0:μ=42000H_0: \mu = 42000 , H1:μ>42000H_1: \mu > 42000 (Claim)
2

Find the critical value

Since α=0.05\alpha = 0.05 and the test is a right-tailed test, the critical value is z = +1.65.
3

Compute the test value

z=xˉ−μσn=43260−42000523030=1.32z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} = \frac{43260 - 42000}{\frac{5230}{\sqrt{30}}} = 1.32
4

Make the decision

Since the test value, +1.32, is less than the critical value, +1.65, and not in the critical region, the decision is “Do not reject the null hypothesis.”
H0H_0 TrueH0H_0 False
Reject H0H_0Type I ErrorCorrect Decision
Do not Reject H0H_0Correct DecisionType II Error
5

Summarize the results.

There is not enough evidence to support the claim that assistant professors earn more on average than $42,000 a year.skjsi
Critical area adalah area antara test value dengan z (1.32 - 1.65)

Two data

Soal : Sebuah survei menemukan bahwa rata-rata tarif kamar hotel di New Orleans adalah $88.42 dan rata-rata tarif kamar di Phoenix adalah $80.61. Asumsikan bahwa data diperoleh dari dua sampel 50 hotel masing-masing dan bahwa standar deviasi adalah $5.62 dan $4.83 masing-masing. Pada α = 0.05, dapatkah disimpulkan bahwa tidak ada perbedaan signifikan dalam tarif?
Solusi :
1

State the hypotheses and identify the claim

H0:μ1=μ2H_0: \mu_1 = \mu_2 (Claim) , H1:μ1≠μ2H_1: \mu_1 \neq \mu_2
2

Find the critical value

Since α=0.05\alpha = 0.05 and the test is a two-tailed test, the critical value is z = +1.96.
3

Compute the test value

z=(xˉ1−xˉ2)−(μ1−μ2)σ12n1+σ22n2=88.42−80.615.62250+4.83250=7.45z = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} = \frac{88.42 - 80.61}{\sqrt{\frac{5.62^2}{50} + \frac{4.83^2}{50}}} = 7.45
4

Make the decision

Reject the null hypothesis at α = 0.05, since 7.45 > 1.96.
H0H_0 TrueH0H_0 False
Reject H0H_0Type I ErrorCorrect Decision
Do not Reject H0H_0Correct DecisionType II Error
5

Summarize the results.

There is enough evidence to reject the claim that the means are equal. Hence, there is a significant difference in the rates.

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P-Values

Soal : Seorang peneliti ingin menguji klaim bahwa rata-rata usia penjaga pantai di Ocean City lebih dari 24 tahun. Dia memilih sampel 36 penjaga dan menemukan rata-rata sampel adalah 24,7 tahun, dengan standar deviasi 2 tahun. Apakah ada bukti untuk mendukung klaim pada α = 0,05? Temukan P-value.
Solusi :
1

State the hypotheses and identify the claim

H0:μ≤24H_0: \mu \leq 24 , H1:μ>24H_1: \mu > 24 (Claim)
2

Compute the test value

z=xˉ−μσn=24.7−24236=2.10z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} = \frac{24.7 - 24}{\frac{2}{\sqrt{36}}} = 2.10
3

find the corresponding area

Using Table E in Appendix C, find the corresponding area under the normal distribution for z = 2.10. It is 0.4821
Biasanya Menggunakan Tabel Ke 2
4

Subtract

Subtract this value for the area from 0.5000 to find the area in the right tail.
    ~~~~0.5000 – 0.4821 = 0.0179
Hence the P-value is 0.0179.
  • apabila Step 1 menghasilkan two tailed test maka hasil P-Value dikalikan 2
5

Make the decision

Since the P-value is less than 0.05, the decision is to reject the null hypothesis.
6

Summarize the results.

There is enough evidence to support the claim that the average age of lifeguards in Ocean City is greater than 24 years.
  • P-Value > α = Not enough evidence to reject
  • P-Value < α = Enough evidence to support

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Small Sample Mean Test

One data

Soal : Sebuah direktur penempatan kerja mengklaim bahwa rata-rata gaji awal perawat adalah $24.000. Sampel 10 perawat memiliki rata-rata $23.450 dan standar deviasi $400. Apakah ada cukup bukti untuk menolak klaim direktur pada α = 0,05?
Solusi :
1

State the hypotheses and identify the claim

H0:μ=24000H_0: \mu = 24000 (Claim) , H1:μ≠24000H_1: \mu \neq 24000
2

Find the critical value

Since α = 0.05 and the test is a two-tailed test, the critical values are t = –2.262 and +2.262 with d.f. = 9.
d.f : degrees of freedom = n-1
3

Compute the test value

t=xˉ−μsn=23450−2400040010=−4.35t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} = \frac{23450 - 24000}{\frac{400}{\sqrt{10}}} = -4.35
4

Reject the null hypothesis

since –4.35 < –2.262.
  • t > critical values = Not enough evidence to support
  • t < critical values = Enough evidence to Reject
5

Summarize the results

There is enough evidence to reject the claim that the starting salary of nurses is $24,000.skjsi

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Proportion Test

np≥5np \geq 5 (Peluang Sukses)
nq≤5nq \leq 5 (Peluang Gagal)

Rumus

z=X−μσz = \frac{X - \mu}{\sigma}, or z=X−npnpqz = \frac{X - np}{\sqrt{npq}}
where, μ=np\mu = np and σ=npq\sigma = \sqrt{npq}

Soal : Seorang pendidik memperkirakan bahwa tingkat putus sekolah untuk senior di sekolah menengah di Ohio adalah 15%. Tahun lalu, 38 senior dari sampel acak 200 senior Ohio menarik diri. Pada α = 0,05, apakah ada cukup bukti untuk menolak klaim pendidik?
Solusi :
1

State the hypotheses and identify the claim

H0:p=0.15H_0: p = 0.15 (Claim) , H1:p≠0.15H_1: p \neq 0.15
2

Find the mean and standard deviation

μ=np=(200)(0.15)=30\mu = np = (200)(0.15) = 30
σ=npq=(200)(0.15)(0.85)=5.05\sigma = \sqrt{npq} = \sqrt{(200)(0.15)(0.85)} = 5.05
3

Find the critical values

Since α = 0.05 and the test is two-tailed the critical values are z=±1.96z = \pm 1.96.
4

Compute the test value

z=X−npnpq=38−305.05=1.58z = \frac{X - np}{\sqrt{npq}} = \frac{38 - 30}{5.05} = 1.58
5

Summarize the results

Do not reject the null hypothesis, since the test value falls outside the critical region.There is not enough evidence to reject the claim that the dropout rate for seniors in high schools in Ohio is 15%.skjsi
For one-tailed test for proportions, follow procedures for the large sample mean test.

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Varian or Standard deviation Test

Rumus

X2=(n−1)s2σ2X^2 = \frac{(n-1)s^2}{\sigma^2}, with d.f. = n-1
where,n = sample size,s2s^2 = sample standard deviation/variance,σ2\sigma^2 = population standard deviation/variance

Soal : Seorang instruktur ingin melihat apakah variasi skor dari 23 siswa di kelasnya lebih kecil dari varians populasi. Varians kelas adalah 198. Apakah ada cukup bukti untuk mendukung klaim bahwa variasi siswa lebih kecil dari varians populasi (α2=α^2 = 225) pada α = 0,05?
Solusi :
1

State the hypotheses and identify the claim

H0:σ2≥225H_0: \sigma^2 \geq 225 (Claim) , H1:σ2<225H_1: \sigma^2 < 225
2

Find the critical value

Since this test is left-tailed and α = 0.05, use the value 1 – 0.05 = 0.95.The d.f. = 23 – 1 = 22.Hence, the critical value is 12.338.
3

Compute the test value

X2=(n−1)s2σ2=(23−1)(198)225=19.36X^2 = \frac{(n-1)s^2}{\sigma^2} = \frac{(23-1)(198)}{225} = 19.36
4

Make a decision

Do not reject the null hypothesis, since the test value falls outside the critical region.skjsi
When the test is two-tailed, you will need to find x2x^2 (left) and x2x^2 (right) and check whether the test value is less than x2x^2 (left) or whether it is greater than x2x^2 (right) in order to reject the null hypothesis.

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The Standard Normal Distribution

Soal : Setiap bulan, sebuah rumah tangga Amerika menghasilkan rata-rata 28 pon kertas koran untuk sampah atau daur ulang. Asumsikan standar deviasi adalah 2 pon. Asumsikan jumlah yang dihasilkan terdistribusi secara normal.Jika rumah tangga dipilih secara acak, temukan probabilitas menghasilkan:
Solusi : a. Lebih dari 30 pon setiap bulan
1

First find the z-value for 30.2

z=x−μσ=30.2−282=1.1z = \frac{x - \mu}{\sigma} = \frac{30.2 - 28}{2} = 1.1
2

Find the area to the right of 30.2

  • Thus, P(z > 1.1) = 0.5 - 0.3643 = 0.1357.
  • itu artinya, probabilitas bahwa rumah tangga yang dipilih secara acak akan menghasilkan lebih dari 30,2 pon koran adalah 0,1357 atau 13,57%.
  • skjsi

Solusi : b. Diantara 27 dan 31 pon setiap bulan
1

First find the z-value for 27 and 31

z1=27−282=−0.5z_1 = \frac{27 - 28}{2} = -0.5
z2=31−282=1.5z_2 = \frac{31 - 28}{2} = 1.5
2

Find the area to the right of 30.2

  • Thus, P(−0.5≤z≤1.5-0.5 \leq z \leq 1.5) = 0.1915 + 0.4332 = 0.6247.
  • itu artinya, probabilitas bahwa rumah tangga yang dipilih secara acak akan menghasilkan antara 27 dan 31 pon koran adalah 0,6247 atau 62,47%.
  • skjsi