> ## Documentation Index
> Fetch the complete documentation index at: https://v1-learn.neoartd.my.id/llms.txt
> Use this file to discover all available pages before exploring further.

# Kontrak Perkuliahan

# Hyspothesis Testing

# Large Sample Mean Test

<Card title="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.

  <br />

  **Solusi** :

  <Steps>
    <Step title="State the hypotheses and identify the claim">
      $H_0: \mu = 42000$ , $H_1: \mu > 42000$ (**Claim**)
    </Step>

    <Step title="Find the critical value">
      Since $\alpha = 0.05$ and the test is a `right-tailed test`, the critical value is z = +1.65.
    </Step>

    <Step title="Compute the test value">
      $z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} = \frac{43260 - 42000}{\frac{5230}{\sqrt{30}}} = 1.32$
    </Step>

    <Step title="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.`”

      <br />

      |                     | $H_0$ True       | $H_0$ False      |
      | ------------------- | ---------------- | ---------------- |
      | Reject $H_0$        | Type I Error     | Correct Decision |
      | Do not Reject $H_0$ | Correct Decision | Type II Error    |
    </Step>

    <Step title="Summarize the results.">
      There is
      not enough evidence to support the
      claim that assistant professors earn
      more on average than \$42,000 a year.

      <img src="https://mintlify.s3.us-west-1.amazonaws.com/mmn/series/kuliah/2/statistika-dasar/images/screenshot-1717503388018.png" alt="skjsi" />

      <Info>Critical area adalah area antara test value dengan z (1.32 - 1.65)</Info>
    </Step>
  </Steps>
</Card>

<Card title="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?

  <br />

  **Solusi** :

  <Steps>
    <Step title="State the hypotheses and identify the claim">
      $H_0: \mu_1 = \mu_2$ (**Claim**) , $H_1: \mu_1 \neq \mu_2$
    </Step>

    <Step title="Find the critical value">
      Since $\alpha = 0.05$ and the test is a `two-tailed test`, the critical value is z = +1.96.
    </Step>

    <Step title="Compute the test value">
      $z = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} = \frac{88.42 - 80.61}{\sqrt{\frac{5.62^2}{50} + \frac{4.83^2}{50}}} = 7.45$
    </Step>

    <Step title="Make the decision">
      Reject the
      null hypothesis at α = 0.05, since
      7.45 > 1.96.

      <br />

      |                     | $H_0$ True       | $H_0$ False      |
      | ------------------- | ---------------- | ---------------- |
      | Reject $H_0$        | Type I Error     | Correct Decision |
      | Do not Reject $H_0$ | Correct Decision | Type II Error    |
    </Step>

    <Step title="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.
    </Step>
  </Steps>
</Card>

## P-Values

<Card>
  **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.

  <br />

  **Solusi** :

  <Steps>
    <Step title="State the hypotheses and identify the claim">
      $H_0: \mu \leq 24$ , $H_1: \mu > 24$ (**Claim**)
    </Step>

    <Step title="Compute the test value">
      $z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} = \frac{24.7 - 24}{\frac{2}{\sqrt{36}}} = 2.10$
    </Step>

    <Step title="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
      <Note>Biasanya Menggunakan Tabel Ke 2</Note>
    </Step>

    <Step title="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.

      <Note>
        * apabila Step 1 menghasilkan **two tailed test** maka hasil P-Value dikalikan 2
      </Note>
    </Step>

    <Step title="Make the decision">
      Since the
      P-value is **less than** 0.05, the decision is
      to `reject the null hypothesis`.
    </Step>

    <Step title="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.

      <Note>
        * P-Value > α = Not enough evidence to reject
        * P-Value \< α = Enough evidence to support
      </Note>
    </Step>
  </Steps>
</Card>

# Small Sample Mean Test

<Card title="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?

  <br />

  **Solusi** :

  <Steps>
    <Step title="State the hypotheses and identify the claim">
      $H_0: \mu = 24000$ (**Claim**) , $H_1: \mu \neq 24000$
    </Step>

    <Step title="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.
      <Note>d.f : degrees of freedom = n-1</Note>
    </Step>

    <Step title="Compute the test value">
      $t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} = \frac{23450 - 24000}{\frac{400}{\sqrt{10}}} = -4.35$
    </Step>

    <Step title="Reject the null hypothesis">
      since –4.35 \< –2.262.

      <Note>
        * t > critical values = Not enough evidence to support
        * t \< critical values = Enough evidence to Reject
      </Note>
    </Step>

    <Step title="Summarize the results">
      There is enough evidence to
      reject the claim that the starting salary
      of nurses is \$24,000.

      <img src="https://mintlify.s3.us-west-1.amazonaws.com/mmn/series/kuliah/2/statistika-dasar/images/screenshot-1717505618217.png" alt="skjsi" />
    </Step>
  </Steps>
</Card>

# Proportion Test

<Note>
  $np \geq 5$ (Peluang Sukses) \
  $nq \leq 5$ (Peluang Gagal)
</Note>

<Card title="Rumus">
  $z = \frac{X - \mu}{\sigma}$, or $z = \frac{X - np}{\sqrt{npq}}$ \
  where, $\mu = np$ and $\sigma = \sqrt{npq}$
</Card>

<Card>
  **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?

  <br />

  **Solusi** :

  <Steps>
    <Step title="State the hypotheses and identify the claim">
      $H_0: p = 0.15$ (**Claim**) , $H_1: p \neq 0.15$
    </Step>

    <Step title="Find the mean and standard deviation">
      $\mu = np = (200)(0.15) = 30$\
      $\sigma = \sqrt{npq} = \sqrt{(200)(0.15)(0.85)} = 5.05$
    </Step>

    <Step title="Find the critical values">
      Since
      α = 0.05 and the test is **two-tailed** the
      critical values are $z = \pm 1.96$.
    </Step>

    <Step title="Compute the test value">
      $z = \frac{X - np}{\sqrt{npq}} = \frac{38 - 30}{5.05} = 1.58$
    </Step>

    <Step title="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%.

      <img src="https://mintlify.s3.us-west-1.amazonaws.com/mmn/series/kuliah/2/statistika-dasar/images/screenshot-1717513201665.png" alt="skjsi" />

      <Note>For one-tailed test for
      proportions, follow procedures for the
      large sample mean test.</Note>
    </Step>
  </Steps>
</Card>

# Varian or Standard deviation Test

<Card title="Rumus">
  $X^2 = \frac{(n-1)s^2}{\sigma^2}$, with d.f. = n-1 \
  where,

  n = sample size,

  $s^2$ = sample standard deviation/variance,

  $\sigma^2$ = population standard deviation/variance
</Card>

<Card>
  **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 ($α^2 = $ 225) pada α = 0,05?

  <br />

  **Solusi** :

  <Steps>
    <Step title="State the hypotheses and identify the claim">
      $H_0: \sigma^2 \geq 225$ (**Claim**) , $H_1: \sigma^2 < 225$
    </Step>

    <Step title="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.
    </Step>

    <Step title="Compute the test value">
      $X^2 = \frac{(n-1)s^2}{\sigma^2} = \frac{(23-1)(198)}{225} = 19.36$
    </Step>

    <Step title="Make a decision">
      `Do not reject the null
            hypothesis, since the test value falls
            outside the critical region.`

      <img src="https://mintlify.s3.us-west-1.amazonaws.com/mmn/series/kuliah/2/statistika-dasar/images/screenshot-1717513793105.png" alt="skjsi" />

      <Note>
        When the test is two-tailed, you
        will need to find $x^2$
        (left) and $x^2$
        (right)
        and check whether the test value is less
        than $x^2$
        (left) or whether it is greater than $x^2$
        (right) in order to reject the null hypothesis.
      </Note>
    </Step>
  </Steps>
</Card>

# The Standard Normal Distribution

<Card>
  **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:

  <br />

  **Solusi** : a. Lebih dari 30 pon setiap bulan

  <Steps>
    <Step title="First find the z-value for 30.2">
      $z = \frac{x - \mu}{\sigma} = \frac{30.2 - 28}{2} = 1.1$
    </Step>

    <Step title="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%.
      *       <img src="https://mintlify.s3.us-west-1.amazonaws.com/mmn/series/kuliah/2/statistika-dasar/images/screenshot-1717515995536.png" alt="skjsi" />
    </Step>
  </Steps>

  <br />

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

  <Steps>
    <Step title="First find the z-value for 27 and 31">
      $z_1 = \frac{27 - 28}{2} = -0.5$\
      $z_2 = \frac{31 - 28}{2} = 1.5$
    </Step>

    <Step title="Find the area to the right of 30.2">
      * Thus, P($-0.5 \leq z \leq 1.5$) = 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%.
      *       <img src="https://mintlify.s3.us-west-1.amazonaws.com/mmn/series/kuliah/2/statistika-dasar/images/screenshot-1717516232241.png" alt="skjsi" />
    </Step>
  </Steps>
</Card>